Question

a) A ball of charge +q is placed on the x-axis at x=0. A second ball of charge +16q is placed on the x-axis at x = -2a. A third ball , with a charge that has a magnitude of q, is placed on the x - axis at a n unknown location . If the net electrostatic force exerted on the ball at the origin due to the other two balls has a magnitude of 5kq^2/a^2, what is the location of the thir d ball and the sign of the third ball

Answer 1

A) The magnitude of the third charge is q and we are assuming that it is poeitively charged and is located at a distance d from the origin.

The force F₁ on the charge +q located at origin due to charge + 16q is

Both the charges are positive so they will repel exch other and the direction of the force is towards right.

The force F₂ on the charge +q located at origin due to charge + q which is located as a distance d from the origin can be calculated as

Directed towards left. As force is a vector quantity and both the forces are directed in the opposite direction so the net force F is

It is given

On doing the simple calculations we get

Third charge can be located at and it is positively charge.

The other possibility can be the charge on third ball has negative sign. In that case the magnitude of the forces F₁ and F₂ will be same but the directions can be different. The direction of F₁ and F₂ will be towards right. Both the forces are acting in the same direction so the resultant force F is

By using the same steps as above

we get

.Among all the possibilities, the location of the third charge d=-a is closest to +16 q charge. So we consider this location for solving the remaining parts of the questions.

B) Electrostatic energy is a scalar quantity and can be calculated as

Here the two charges q₁ and q₂ are separated by a distance r. We have three charges so we need to consider the interactions among all the charges which give rise to three terms for electrostatic energy

The electrostatic energy is

.

C) The electric potential V due to the charge q at location r is

V is a scalar quantity so the net potential at x=+2a due to all the charges can be written as

D} Electric field E₁ due to charge +q at x=+2a is

Directed towards right.

Electric field E₂ due to charge -q at x=+2a is

Directed towards left.

Electric field E₃ due to charge +16q at x=+2a is

Directed towards right. Since electric field is a vector quantity so we need to consider the directions as well. The resultant field E is given at

The resultant field is directed towards right.