A small sphere has a charge of q = 5e, where e is the charge on an electron.(a) Determine the electric potential (in V) due to the charge at a distance r = 0.300 cm from the charge.v(b) Determine the electric potential difference (in V) between a point that is 5r away and this point, that is V(5r) – V(r).v(c) Determine the electric potential difference (in V) between a point that is - away and this point, that is VV

a) Given data The magnitude of the charge on the sphere is given as q = 5e. The distance of the…

Answered: A small sphere has a charge of q = 5e,…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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A small sphere has a charge of q = 5e, where e is the charge on an electron.
(a) Determine the electric potential (in V) due to the charge at a distance r = 0.300 cm from the charge.
v
(b) Determine the electric potential difference (in V) between a point that is 5r away and this point, that is V(5r) – V(r).
v
(c) Determine the electric potential difference (in V) between a point that is - away and this point, that is V
V

Expert Solution

Step 1

a) Given data

The magnitude of the charge on the sphere is given as q = 5e.
The distance of the point from the sphere is given as r = 0.3 cm.
1e = -1.6 x 10^-19 C.

The expression for the electric potential is given as,

$V (r) = \frac{k q}{r}$

Here, k is the coulomb's constant.

On putting the values as,

$V (r) = \frac{k q}{r} V (r) = \frac{9 \times 10^{9} {Nm}^{2} C^{- 2} \times (5 e)}{0.3 \times 10^{- 2} m} V (r) = \frac{9 \times 10^{9} {Nm}^{2} C^{- 2} \times (5 \times - 1.6 \times 10^{- 19} C)}{0.3 \times 10^{- 2} m} V (r) = - 2.4 \times 10^{- 6} V$

Thus, the potential at a distance 0.260 cm away is -2.4 x 10^-6 V.

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