Question

A solid insulating sphere has total charge Q and radius R. The sphere's charge is distributed uniformly throughout its volume. Let r be the radial distance measured from the center of the sphere. If E = 520 N/C at r= R/4, what is E at r = 4R?

Answer 1

Answer:

2.03N/C

Explanation:

Electric field, E, is given as:

E = kQ/r²

When r = R/4, E = 520N/C.

=> 520 = kQ/(R/4)²

520 = 16kQ/R²

=> R² = 16kQ/520 ________ (1)

When r = 4R,

E = kQ/(4R)²

E = kQ/16R²

=> R² = kQ/16E __________ (2)

Equating (1) & (2):

16kQ/520 = kQ/16E

=> E = 520/(16 * 16)

E = 520/256

E = 2.03 N/C

Answer 2

The electric field at a radial distance r = 4R  of the sphere is 130 N/C.

The volume of a sphere is given as;

$V = \frac{4}{3} \pi R^3$

The enclosed charge of the sphere is calculated as follows;

$q _{in} = \frac{Q}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi (\frac{R}{4} )^3\\\\ q_{in} = \frac{Q}{64}$

The electric flux is calculated as follows;

$EA = \frac{q_{in}}{\varepsilon _0}$

where;

• E is the electric field
• A is the area of the sphere

$E \times 4\pi r^2 = \frac{Q}{64\varepsilon _0} \\\\ E \times 4\pi \times (\frac{R}{4} )^2 = \frac{Q}{64\varepsilon _0} \\\\ E\times 4\pi R^2 = \frac{16Q}{64 \varepsilon_0} \\\\ E\times 4\pi R^2 = \frac{Q}{4\varepsilon _0} \\\\ E \times 4R^2 = \frac{Q}{4 \pi\varepsilon _0}\\\\ E \times 4R^2 = kQ\\\\ E = \frac{kQ}{4R^2} \\\\ 4E = \frac{kQ}{R^2} \\\\ 4(520) = \frac{kQ}{R^2}$

The electric field at a radial distance r = 4R  of the sphere is calculated as;

$E = \frac{kQ}{r^2} \\\\ E_2 = \frac{kQ}{(4R)^2} \\\\ E_2 = \frac{kQ}{16R^2} \\\\ E_2= \frac{1}{16} (\frac{kQ}{R^2} )\\\\ E_2 = \frac{1}{16} \times E\\\\ E_2 = \frac{1}{16} \times 4(520)\\\\ E_2 = 130 \ N/C$

The electric field at a radial distance r = 4R  of the sphere is 130 N/C.

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