Question

An insulating sphere of radius 13 cm has a uniform charge density throughout its volume. 13 cm 21.6 cm 7.5 cm p If the magnitude of the electric field at a distance of 7.5 cm from the center is 78400 N/C , what is the magnitude of the electric field at 21.6 cm from the center? Answer in units of N/C

Answer 1

Answer:

The value is $E_1 = 49224.1 \ N/C$

Explanation:

From the question we are told that

The radius is $r = 13 \ cm = 0.13 \ m$

The electric field is $E = 78400 \ N/C$ at a distance $d = 7.5 \ cm = 0.075 \ m$

Generally the electric field at a distance $d = 7.5 \ cm = 0.075 \ m$ is mathematically represented as

$E = \frac{k * q * d}{r^3}$ Note: the reason we are using this

formula is because $d < r$

Here k is the coulomb constant with value $k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $78400= \frac{9*10^{9} * q * 0.075}{0.13^3 }$

Generally the electric field at a distance $d = 21.6 \ cm = 0.216 \ m$ is mathematically represented as

$E_1 = \frac{k * q }{^2}$ Note: the reason we are using this

formula is because $d > r$

Now dividing $E_1\ \ by \ \ E$

$\frac{E_1}{78400} = \frac{\frac{9*10^9 * q}{0.216^2} }{\frac{9*10^9 * q * 0.075}{0.13^3} }$

=> $E_1 = \frac{0.13 ^3}{ 0.075 * 0.216^2} * 78400$

=> $E_1 = 49224.1 \ J$