Question

A.How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1100 N/C just outside the surface of the sphere?

N=?

B.What is the electric field at a point 12.5 cm outside the surface of the sphere?

E=?

Answer 1

Solution:-

1. Calculate the excess electron

According to gauss law

Gauss law : EA = Q/ε₀

=> Q = EAε₀

Sphere surface area : A = 4πr²:

=> Q = E4πε₀r²

      Q = Er²/k

(Calculate either with ε₀ = 8.85×10 ̄¹² F/m or k = 8.99×10⁹ Nm²/C²)

=> Q = 1100N/C*(12.5m)² / 8.99×10⁹Nm²/C²

Q = 1.9×10¹³ C

(1.9*10¹³ C) / (1.6*10 ̄ ¹⁹C/electron)

= 1.18*10^-6 = electrons in excess.

2)First calculate the value of charge ,so by using the formula of electric field,

Electric field :-

An electric field (sometimes abbreviated as E-field) is a vector field surrounding an electric charge that exerts force on other charges, attracting or repelling them.[1] [2] Mathematically the electric field is a vector field that associates to each point in space the force, called the Coulomb force, that would be experienced per unit of charge, by an infinitesimal test charge at that point.[3][4][5] The units of the electric field in the SI system are newtons per coulomb (N/C), or volts per meter (V/m). Electric fields are created by electric charges, and by time-varying magnetic fields.

E = k*q / r^2

=> 1100 = (9x10^9)(q) / (12.5)^2

=> q = 1.90*10¹³ C

A point 12.5 cm outside the surface of the sphere is 25.0 cm from the center, therefore calculate electric field

E = (9x10^9)(1.90*10¹³) / (0.025)^2

E = 2.63*10²⁶ N/C

Elcetric field outside the sphere is =  E = 2.63*10²⁶ N/C