Question

The sides of a rectangle abcd are 15cm and 5 cm. point charge of -5 micro coulomb and +2 micro coulomb are placed at vertices b and d respectively. calculate the electric potential at vertices a and c. also calculate the work done in carrying a charge of +3 micro coulomb from c to a.​

Answer 1

Given:

The sides of a rectangle abcd are 15cm and 5 cm.

A point charge of -5 micro coulomb and +2 micro coulomb are placed at vertices b and d respectively.

To find:

Calculate the electric potential at vertices a and c. Also calculate the work done in carrying a charge of +3 micro coulomb from c to a.​

Solution:

The electrical potential at vertex a is given as follows:

$V_a=k(\dfrac{-5 \times 10^{-6}}{5 \times 10^{-2}} +\dfrac{2 \times 10^{-6}}{15 \times 10^{-2}})\\\\= 9 \times 10^9(10^{-4}+\dfrac{2}{15}\times 10^{-4})\\\\=-\dfrac{13}{15} \times 10^{-4} \times 9 \times 10^9\\\\\therefore V_a=-7.8 \times 10^5 V$

The electrical potential at vertex c is given as follows:

$V_c=k(\dfrac{-5 \times 10^{-6}}{15 \times 10^{-2}} +\dfrac{2 \times 10^{-6}}{5 \times 10^{-2}})\\\\= 9 \times 10^9(-\dfrac{1}{3}+\dfrac{2}{5})\times 10^{-4}\\\\=-\dfrac{1}{15} \times 10^{-4} \times 9 \times 10^9\\\\\therefore V_c=0.3 \times 10^5 V$

The work done in carrying a charge of +3 micro coulomb from c to a is given as follows:

We use the formula,

W = q (Vc - Va)

= 3 × 10^{-6} (0.3 × 10^5 + 7.8 × 10^5)

= 3 × 10^{-6} (8.1 × 10^5)

= 2.43 J

∴ W = 2.43 J