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

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1