Given:

• L = 0.5 m
• λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
• d = 0.2 m
• k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

Given:

• L = 0.5 m
• λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
• d = 0.2 m
• k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

Given:

• L = 0.5 m
• λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
• d = 0.2 m
• k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

Given:

• L = 0.5 m
• λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
• d = 0.2 m
• k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

Given:

• L = 0.5 m
• λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
• d = 0.2 m
• k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

Given:

• L = 0.5 m
• λ(x) = λ₀x where λ₀ = 4.0 × 10⁻⁶ C/m²
• d = 0.2 m
• k = 8.99 × 10⁹ N·m²/C²

Setup:

For a small element dx at position x from the origin, the charge is: $dq = λ(x)dx = λ_0 x \, dx$

The electric field contribution at P (located at distance d from x = 0): $dE = k\frac{dq}{r^2} = k\frac{λ_0 x \, dx}{(d + x)^2}$

Integration:

$E = \int_0^L k\frac{λ_0 x}{(d + x)^2} \, dx = kλ_0 \int_0^L \frac{x}{(d + x)^2} \, dx$

Using substitution u = d + x, du = dx, when x = 0, u = d; when x = L, u = d + L: $E = kλ_0 \int_d^{d+L} \frac{u - d}{u^2} \, du = kλ_0 \int_d^{d+L} \left(\frac{1}{u} - \frac{d}{u^2}\right) du$

$E = kλ_0 \left[\ln u + \frac{d}{u}\right]_d^{d+L}$

$E = kλ_0 \left[\ln(d+L) - \ln d + \frac{d}{d+L} - 1\right]$

$E = kλ_0 \left[\ln\left(\frac{d+L}{d}\right) + \frac{d}{d+L} - 1\right]$

Numerical calculation: $E = (8.99 \times 10^9)(4.0 \times 10^{-6})\left[\ln\left(\frac{0.7}{0.2}\right) + \frac{0.2}{0.7} - 1\right]$

$E = 35,960 \left[\ln(3.5) + 0.286 - 1\right]$

$E = 35,960 \left[1.253 + 0.286 - 1\right] = 35,960(0.539)$

$E = 1.94 \times 10^4 \text{ N/C}$

Answer: The electric field at point P is 1.94 × 10⁴ N/C pointing away from the rod.

Given:

• R = 0.1 m
• Q = 2.0 × 10⁻⁸ C = 20 nC
• k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

$\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}$

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

Given:

• R = 0.1 m
• Q = 2.0 × 10⁻⁸ C = 20 nC
• k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

$\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}$

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

Given:

• R = 0.1 m
• Q = 2.0 × 10⁻⁸ C = 20 nC
• k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

$\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}$

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

Given:

• R = 0.1 m
• Q = 2.0 × 10⁻⁸ C = 20 nC
• k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

$\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}$

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

Given:

• R = 0.1 m
• Q = 2.0 × 10⁻⁸ C = 20 nC
• k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

$\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}$

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

Given:

• R = 0.1 m
• Q = 2.0 × 10⁻⁸ C = 20 nC
• k = 8.99 × 10⁹ N·m²/C²

Formula for field on axis: $E_x = \frac{kQx}{(x^2 + R^2)^{3/2}}$

(a) Field at x = 0.05 m:

$E = \frac{(8.99 \times 10^9)(2.0 \times 10^{-8})(0.05)}{[(0.05)^2 + (0.1)^2]^{3/2}}$

$E = \frac{(8.99)(0.1)}{[(0.0025 + 0.01)]^{3/2}} = \frac{0.899}{(0.0125)^{3/2}}$

$E = \frac{0.899}{1.40 \times 10^{-3}} = 6.42 \times 10^2 \text{ N/C}$

(b) Maximum field position:

To find maximum, take derivative and set equal to zero: $\frac{dE_x}{dx} = kQ\frac{d}{dx}\left[\frac{x}{(x^2 + R^2)^{3/2}}\right] = 0$

$\frac{dE_x}{dx} = kQ\frac{(x^2 + R^2)^{3/2} - x \cdot \frac{3}{2}(x^2 + R^2)^{1/2}(2x)}{(x^2 + R^2)^3}$

$= kQ\frac{(x^2 + R^2) - 3x^2}{(x^2 + R^2)^{5/2}} = 0$

$R^2 - 2x^2 = 0$

$x = \frac{R}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.0707 \text{ m}$

(c) Maximum field strength:

$E_{max} = \frac{kQ(R/\sqrt{2})}{[(R/\sqrt{2})^2 + R^2]^{3/2}}$

$= \frac{kQ(R/\sqrt{2})}{[R^2/2 + R^2]^{3/2}} = \frac{kQ(R/\sqrt{2})}{(3R^2/2)^{3/2}}$

$= \frac{kQ}{R^2\sqrt{2}(3/2)^{3/2}} = \frac{kQ}{R^2\sqrt{2} \cdot \frac{3\sqrt{3}}{2\sqrt{2}}}$

$= \frac{2kQ}{3\sqrt{3}R^2} = \frac{2(8.99 \times 10^9)(2.0 \times 10^{-8})}{3\sqrt{3}(0.1)^2}$

$= \frac{359.6}{0.052} = 6.91 \times 10^2 \text{ N/C}$

Answers: (a) E = 642 N/C (b) x = 7.07 cm = R/√2 (c) E_max = 691 N/C

Given:

• d = 2.0 × 10⁻³ m = 2.0 mm
• p = 5.0 × 10⁻¹² C·m
• r = 0.1 m = 10 cm
• k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

$q = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}$

(b) Field on perpendicular bisector (far field, r >> d):

$E_{perp} = \frac{kp}{r^3}$

$E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}$

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

$E_{axis} = \frac{2kp}{r^3}$

$E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}$

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: $E_{axis} = 2E_{perp}$

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

Given:

• d = 2.0 × 10⁻³ m = 2.0 mm
• p = 5.0 × 10⁻¹² C·m
• r = 0.1 m = 10 cm
• k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

$q = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}$

(b) Field on perpendicular bisector (far field, r >> d):

$E_{perp} = \frac{kp}{r^3}$

$E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}$

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

$E_{axis} = \frac{2kp}{r^3}$

$E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}$

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: $E_{axis} = 2E_{perp}$

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

Given:

• d = 2.0 × 10⁻³ m = 2.0 mm
• p = 5.0 × 10⁻¹² C·m
• r = 0.1 m = 10 cm
• k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

$q = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}$

(b) Field on perpendicular bisector (far field, r >> d):

$E_{perp} = \frac{kp}{r^3}$

$E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}$

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

$E_{axis} = \frac{2kp}{r^3}$

$E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}$

Direction: Away from dipole (from -q toward +q) for points on the side of +q

Note: The axial field is exactly twice the perpendicular field at the same distance: $E_{axis} = 2E_{perp}$

Answers: (a) q = 2.5 nC (b) E = 45.0 N/C (perpendicular to axis) (c) E = 89.9 N/C (along axis)**

Given:

• d = 2.0 × 10⁻³ m = 2.0 mm
• p = 5.0 × 10⁻¹² C·m
• r = 0.1 m = 10 cm
• k = 8.99 × 10⁹ N·m²/C²

(a) Magnitude of each charge:

Dipole moment: p = qd

$q = \frac{p}{d} = \frac{5.0 \times 10^{-12}}{2.0 \times 10^{-3}} = 2.5 \times 10^{-9} \text{ C} = 2.5 \text{ nC}$

(b) Field on perpendicular bisector (far field, r >> d):

$E_{perp} = \frac{kp}{r^3}$

$E = \frac{(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{4.495 \times 10^{-2}}{10^{-3}} = 44.95 \text{ N/C}$

Direction: From +q toward -q (perpendicular to dipole axis)

(c) Field on axis (far field, r >> d):

$E_{axis} = \frac{2kp}{r^3}$

$E = \frac{2(8.99 \times 10^9)(5.0 \times 10^{-12})}{(0.1)^3}$

$E = \frac{8.99 \times 10^{-2}}{10^{-3}} = 89.9 \text{ N/C}$

Direction: Away from dipole (from -q toward +q) for points on the side of +q