Question

find the potential at r1=40cm and r2=10cm from a charge Q=2 x 10^-4 and also the potential difference between these two points.

Answer 1

Answer on Question #63106-Physics-Electromagnetism

Find the potential at $r_1 = 40\,\mathrm{cm}$ and $r_2 = 10\,\mathrm{cm}$ from a charge $Q = 2 \times 10^{-4}$ and also the potential difference between these two points.

Solution

The potential at distance $r$ is

$$V(r) = k \frac{Q}{r}$$ $$V_1 = k \frac{Q}{r_1} = (9 \cdot 10^9) \frac{(2 \cdot 10^{-4})}{0.4} = 4.5 \cdot 10^6 V$$ $$V_2 = k \frac{Q}{r_2} = (9 \cdot 10^9) \frac{(2 \cdot 10^{-4})}{0.1} = 18 \cdot 10^6 V$$

The potential difference between these two points is

$$V_2 - V_1 = 18 \cdot 10^6 - 4.5 \cdot 10^6 = 13.5 \cdot 10^6 V.$$

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