Electrostatic Potential due to a Point Charge - Free MCQ Practice Test

Force on a unit point charge kept at a distance r from the charge = q/r². 

Therefore, work done to bring that point charge through a small distance dr = q/r² * (-dr). Therefore, the potential of that point is = .

Charge: q=10 C

Distance: r = 8 cm = 0.08 m

Coulomb constant: k = 9 × 10⁹

^Formula: 

Work done = potential*charge by definition. We know that the potential of a point is the amount of work done to bring a unit charge from infinity to a certain point. Therefore, work done W=q*V=4*10^-3*200J=0.8J. The work done is positive in this case.

Electric potential is a scalar quantity and its value is solely dependent on the charge near it and the distance from that charge. In this case, the point is equidistant from the two point charges and the point charges have the same value but opposite nature. Therefore equal but opposite potentials are generated due to the charges and hence the net potential at midpoint becomes zero.

Electric potential between +q and +q isand electric potential between two charges +q and –q is  . Therefore the electric potential of the system is Electric potential is a scalar quantity and thus the system becomes zero-potential-system.

We know that electric potential does not depend on the path, i.e. it is a state function. Therefore if we rotate a charge around another in a complete circular path, the potential energy of its initial and final points is the same. Therefore, there is no change of potential energy of the charge and hence net work done on the charge is zero.

Increase in potential energy of the electron when it strikes another plate = Potential difference*charge of electron=20*1.6*10^-19 J=3.201*10^-18J and as the electron was at rest initially, this energy will be converted to kinetic energy completely. Therefore, 0.5*mass of electron*(velocity)²=3.201*10^-18J. Substituting m=9.11*10^-31kg, we get v=2.65*10⁶ m/s.

Electric field due to a charge q at a distance r is  and electric potential at that point is  Therefore,  and  Dividing these two equations, we get r=0.5m. If the medium is air, k=1. Thus we can get the value of q=0.89 C.

Let the distance between any two charges is d. Therefore the net potential energy of the system is But the total energy of the system is zero. So, -qQ-qQ+ q²/2 

=0 that means q=4Q, i.e. q/Q=4.

Electric potential is a positive quantity if both the charges are positive. In this case, all the three charges are positive; hence there is no negative term in the total energy term. Therefore, we cannot make the total energy of the system zero by adjusting the values of charge. But it would be possible if one of the charges were positive.