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 0.48 m from it. Find the ratio qB/qA of the charges.

qa/qb = 0.3125

Explanation:

Let the distance of the point from first charge (qa) be ra.

Likewise, let the distance of the point from the second charge (qb) be ra

Now, from the question, ra=0.15m

While rb = 0.48m

Normally, we know that:

The electric potential due to a point charge, q, at a point located at a distance, r, away from it is given by the equation;

V = q/(4π(ϵo)r)

We know that 1/(4π(ϵo)) cam be said to ne K.

Therefore, V = Kq/r

Where K = 9 × 10^(9) V.m/C

Now, since from the question, the electric potential at the point is the same due to each of the charges, their electric potential will be the same, thus;

Va = Vb

So, (Kqa) / ra = (Kqb) / rb

This gives us; qa / ra = qb / rb

So rearranging, we get;

qa/qb = ra/rb = 0.15/0.48 = 0.3125

15.3 cos 65 degrees trig