Question

Q9. (a) What is the potential between two points situated 10 cm and 20 cm from a 3.0 μC point charge?

(b)To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?

Answer 1

The potential difference between points from a point charge is expressed by the equation

$V_A -V_B = kQ(\frac{1}{r_a} - \frac{1}{r_b})$

Constant $k = 9 \times 10^9 \; N \; m^2/C^2$

r_a is position of first point

r_b is position of second point

(a)

$V_A -V_B = (9 \times 10^9 \;Nm^2/C^2)(3.0 \; μC)(\frac{1\;C}{10^6 \;μC})(\frac{1}{0.1 \;m} - \frac{1}{0.2 \;m}) \\ = 135 \times 10^3 \;V$

The potential between two points situated 10 cm and 20 cm from a 3.0 μC point charge is $135 \times 10^3 \;V$ .

(b)

$\frac{1}{r_b}=\frac{1}{r_a}- \frac{V}{kQ} \\ \frac{1}{r_b}=\frac{1}{0.1 \;m}- \frac{2 \times 135 \times 10^3 \;V}{(9 \times 10^9 \;Nm^2/C^2)(3.0 \times 10^{-6}\;C)} \\ \frac{1}{r_b}= 0 \\ r_b = \infin$

Therefore, to double the potential between the points, the 20 cm point has to be shift to infinity.