Question

Two charges A and B are fixed in place, at different distances from a certain spot. At this spot the potentials due to the two charges are equal. Charge A is 0.15 m from the spot, while charge B is 0.50 m from it. Find the ratio qB/qA of the charges.

Answer 1

Answer:

$\frac{q_B}{q_A}=\frac{0.15}{0.5}=0.3$

Explanation:

We are given that two charges A  and B

Distance of charge A from the spot=$r_A=0.15 m$

Distance of charge B from the spot=$r_B=0.50 m$

Potential due to charge A and charge B are equal at the spot.

We have to find the ratio of $\frac{q_B}{q_A}$

According to question

$V_A=V_B$

We know that potential at a point

$V=\frac{Kq}{r}$

Using the formula

$\frac{Kq_A}{0.15}=\frac{Kq_B}{0.5}$

$\frac{q_B}{q_A}=\frac{0.15}{0.5}=0.3$