Question

An isolated charged conducting sphere of radius 15.0 cm creates an electric field of 4.90 x 104 N/C at a distance 22.0 cm from its center (a) What is its surface charge density? Your response is off by a multiple of ten. HC/m2 (b) What is its capacitance? pF Need Help?Resd t

Answer 1

a)

Since, electric field due to conducting sphere = E = k*q/r^2

where, k = 9*10^9

E = 4.90*10^4 N/C

q = charge on sphere = ??

r = 22.0 cm = 0.22 m

So,

q = (E*r^2)/k

q = (4.90*10^4)*(0.22^2)/(9*10^9)

q = 2.63*10^-7 C

Now , surface charge density($\sigma$) = q/A

where, A = surface area = 4*pi*R^2

here R = radius of sphere = 15.0 cm = 0.15 m

So, A = 4*pi*(0.15^2)

A = 0.282 m^2

So,

$\sigma$ = (2.63*10^-7 C)/(0.282 m^2)

$\sigma$ = 9.33*10^-7

$\sigma$ = 0.933 $\mu$C/m^2

(b.)

Since Capacitance of conducting sphere(C) = R/k

So,

C = 0.15/(9*10^9)

C = 1.67*10^-11 F

C = 0.167 pF

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