Question

Question 5 of 7
Consider a charged conducting sphere of radius 25 cm. If the charge on the sphere is 7.2 nC, the potential on its surface is
OA. 259 V
B.320 V
C. 1037 V
D. 1045 V​

Answer 1

Answer:

A. 259 V

Explanation:

Radius of sphere: R = 25 cm

Charge on the sphere: q = 7.2 nC

Potential on surface of sphere:

$\rm V_{surface} = \dfrac{kq}{R} \\ \\ \rm = \dfrac{9 \times {10}^{9} \times 7.2 \times {10}^{ - 9} }{25 \times {10}^{ - 2} } \\ \\ \rm = \frac{64.8}{25} \times {10}^{2} \\ \\ \rm = 2.592 \times {10}^{2} \\ \\ \rm = 259.2 \: V$

Answer 2

Answer: A. 259 V.

Question:

Here,

r = 25 cm or 25 × 10^-2 m

Q = 7.2 nC or, 7.2 × 10^-9 C

[1 C = 10^9 nC]

We know,

Electric potential of a charge at a distance r is = kQ/r

Now,

E.P. = (9 × 10^9 NC^2/m^2)(7.2 × 10^-9 C)/(25 × 10^2 m)

=> EP = (9 × 7.2)/(25 × 10^-2) V

=> EP = 648/25 × 10 V

=> EP = 25.92 × 10 V ≈ 259 V.