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. (b) Determine the electric potential difference (in V) between a point that is 5r away and this point, that is V(5r) – V(r). (c) Determine the electric potential difference (in V) between a point that is - away and this point, that is V=) - vr). V

Principles of Physics: A Calculus-Based Text
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Chapter20: Electric Potential And Capacitance
Section: Chapter Questions
Problem 17P: Two particles each with charge +2.00 C are located on the x axis. One is at x = 1.00 m, and the...
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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
Transcribed Image Text: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

  1. The magnitude of the charge on the sphere is given as q = 5e.
  2. The distance of the point from the sphere is given as r = 0.3 cm.
  3. 1e = -1.6 x 10-19 C.

The expression for the electric potential is given as,

Vr=kqr

Here, k is the coulomb's constant.

On putting the values as,

Vr=kqrVr=9×109 Nm2C-2×5e0.3×10-2 mVr=9×109 Nm2C-2×5×-1.6×10-19 C0.3×10-2 m   Vr=-2.4×10-6 V

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

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