This set of Physics Online Test for Schools focuses on “Electrostatic Potential due to a Point Charge”.

1. Electric potential due to a point charge q at a distance r from the point is _______ (in the air).

a) \(\frac {q}{r}\)

b) q*r

c) \(\frac {q}{r^2}\)

d) \(\frac {-q}{r}\)

View Answer

Explanation: Force on a unit point charge kept at a distance r from the charge=\(\frac {q}{r^2}\). Therefore, work done to bring that point charge through a small distance dr=\(\frac {q}{r^2}\)*(-dr). Therefore, the potential of that point is =\(\int_\alpha^r\frac {-q}{r^2}dr = \frac{q}{r}\).

2. Calculate electric potential due to a point charge of 10C at a distance of 8cm away from the charge.

a) 1.125*10^{13}V

b) 1.125*10^{12}V

c) 2.25*10^{13}V

d) 0.62*10^{13}V

View Answer

Explanation: In the SI system, electric potential due to a point charge at a distance r is \(\frac {q}{4\pi \varepsilon r}\)=9*10

^{9}*\(\frac {q}{r}\). Substituting the values, we get potential=9*10

^{9}*\(\frac {10}{0.08}\)V=1.125*10

^{12}V. Though in practice, this huge value of electric potential is not present.

3. What is the amount of work done to bring a charge of 4*10^{-3}C charge from infinity to a point whose electric potential is 2*10^{2}V?

a) 0.8J

b) -0.8J

c) 1.6J

d) -0.4J

View Answer

Explanation: 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.

4. Two point charges 10C and -10C are placed at a certain distance. What is the electric potential of their midpoint?

a) Some positive value

b) Some negative value

c) Zero

d) Depends on medium

View Answer

Explanation: 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.

5. What is the electric potential of the system?

a) \(\frac {q^2}{4\pi \varepsilon r}\)

b) \(\frac {3q^2}{4\pi \varepsilon r}\)

c) Zero

d) \(\frac {q^2}{4\pi \varepsilon r} * \frac {1}{3}\)

View Answer

Explanation: Electric potential between +q and +q is \(\frac {q^2}{4\pi \varepsilon r}\) and electric potential between two charges +q and –q is \(\frac {-q^2}{4\pi \varepsilon (2r)}\). Therefore the electric potential of the system is \(\frac {1}{4\pi \varepsilon }(\frac {q^2}{r}-\frac {q^2}{2r}-\frac {q^2}{2r})\)=0. Electric potential is a scalar quantity and thus the system becomes zero-potential-system.

6. A small charge q is rotated in a complete circular path of radius r surrounding another charge Q. The work done in this process is _________

a) Zero

b) \(\frac {qQ}{4\pi \varepsilon r}\)

c)\(\frac {q}{4\pi \varepsilon (\pi r)}\)

d) \(\frac {q}{4\pi \varepsilon (2\pi r)}\)

View Answer

Explanation: 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.

7. Two plates are kept at a distance of 0.1m and their potential difference is 20V. An electron is kept at rest on the surface of the plate with lower potential. What will be the velocity of the electron when it strikes another plate?

a) 1.87*10^{6} m/s

b) 2.65*10^{6} m/s

c) 7.02*10^{12} m/s

d) 32*10^{-19} m/s

View Answer

Explanation: 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

^{-18}J 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

^{-18}J. Substituting m=9.11*10

^{-31}kg, we get v=2.65*10

^{6}m/s.

8. Electric field intensity and electric potential at a certain distance from a point charge is 32 N/C and 16 J/C. What is the distance from the charge?

a) 50 m

b) 0.5 m

c) 10 m

d) 7 m

View Answer

Explanation: Electric field due to a charge q at a distance r is \(\frac {q}{kr^2}\) and electric potential at that point is \(\frac {q}{kr}\). Therefore, \(\frac {q}{kr^2}\)=32 and \(\frac {q}{kr}\)=16. 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.

9. Three charges –q, Q and –q are placed in a straight line maintaining equal distance from each other. What should be the ratio \(\frac {q}{Q}\) so that the net electric potential of the system is zero?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: Let the distance between any two charges is d. Therefore the net potential energy of the system is\(\frac {-q*Q}{d}+\frac {-q*Q}{d}+\frac {-q*q}{2d}\). But the total energy of the system is zero. So, -qQ-qQ+\(\frac {q^2}{2}\)=0 that means q=4Q, i.e. \(\frac {q}{Q}\)=4.

10. Three positive charges are kept at the vertices of an equilateral triangle. We can make the potential energy of the system zero by adjusting the amount of charges. This statement is ______

a) True

b) False

View Answer

Explanation: 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.

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