Question

Positive charge Q is distributed uniformly throughout an insulating sphere of radius R, centered at the origin. A particle with charge Q is placed at x = 2R on the x axis. The magnitude of the electric field at x = R/2 on the x-axis is

Answer 1

Answer:

$E=\dfrac{Q}{72\pi \varepsilon _0R^2}$

Explanation:

 Given that

Charge on the sphere = Q

Radius = R

The charge on the particle = Q

x= 2 R

We have to find out electric filed at x= R/2

The electric field due to charge on the sphere :

$4\pi r^2\ E_1=\dfrac{Qr^3}{R^3\varepsilon _0}\\E_1=\dfrac{Qr}{4\pi R^3\varepsilon _0}\\r=\dfrac{R}{2}\\E_1=\dfrac{QR}{8\pi R^3\varepsilon _0}\\E_1=\dfrac{Q}{8\pi R^2\varepsilon _0}$

E₁ is in positive x direction .

The electric field due to point charge :

$E_2=\dfrac{Q}{4\pi \varepsilon _0\left ( 2R-\dfrac{R}{2} \right )^2}\\E_2=\dfrac{Q}{4\pi \varepsilon _0\left ( \dfrac{3R}{2} \right )^2}\\E_2=\dfrac{Q}{9\pi \varepsilon _0R^2}$

E₂ is in negative direction.

Total electric filed E

E=E₁ - E₂

$E=\dfrac{Q}{8\pi \varepsilon _0R^2}-\dfrac{Q}{9\pi \varepsilon _0R^2}\\\\E=\dfrac{Q}{72\pi \varepsilon _0R^2}$

$E=\dfrac{Q}{72\pi \varepsilon _0R^2}$