Physics Questions and Answers – Electrostatic Potential due to a System of Charges

«
»

This set of Physics Question Papers for Schools focuses on “Electrostatic Potential due to a System of Charges”.

1. Which is the expression for electric potential due to a system of charges from the following?
a) V=\(\frac {1}{4 \pi \varepsilon_o} (\frac {q_1}{r_1} + \frac {q_2}{r_2} + \frac {q_3}{r_3} +……. \frac {q_n}{r_n})\)
b) V=1 χ 4πεo \( (\frac {q_1}{r_1} + \frac {q_2}{r_2} + \frac {q_3}{r_3} +……. \frac {q_n}{r_n})\)
c) V=\(\frac {1}{8 \pi \varepsilon_o} (\frac {q_1}{r_1} + \frac {q_2}{r_2} + \frac {q_3}{r_3} +……. \frac {q_n}{r_n})\)
d) V=\(\frac {1}{4 \pi \varepsilon_o} (\frac {q_1}{r_1} \times \frac {q_2}{r_2} \times \frac {q_3}{r_3} \times……. \frac {q_n}{r_n})\)
View Answer

Answer: a
Explanation: The electric potential at a point due to a system of charges is equal to the algebraic sum of the electric potentials due to individual charges at that point. The expression for electric potential due to a system of charges is given by:
V=\(\frac {1}{4 \pi \varepsilon_o} (\frac {q_1}{r_1} + \frac {q_2}{r_2} + \frac {q_3}{r_3} +……. \frac {q_n}{r_n})\)
advertisement

2. There are infinite number of charges, each equal to ‘q’, which are placed along the X-axis at points x = 1, x = 4, x = 16, x = 64……..Then, determine the electric potential due to this system of charges at the point x = 0.
a) V=\(\frac {16q}{3\pi \varepsilon_o}\)
b) V=\(\frac {3q}{4\pi \varepsilon_o}\)
c) V=\(\frac {4q}{8\pi \varepsilon_o}\)
d) V=\(\frac {3q}{16\pi \varepsilon_o}\)
View Answer

Answer: d
Explanation: We know electric potential(V) due to charge ‘q’ is V = \(\frac {kq}{r}\).
Electric potential (V) is a scalar quantity, so, the total potential at x = 0 is the sum of all the individual charges
V=[\((\frac {kq}{1}) + (\frac {kq}{4}) + (\frac {kq}{16}) + (\frac {kq}{64}) \) + ……..]
V=kq\((\frac {1}{(1-\frac{1}{4})})\) → sum of infinite G.P=\(\frac {a}{(1-r)}\)
V=\((\frac {q}{4\pi \varepsilon_o}) \times (\frac {3}{4})\)
V=\(\frac {3q}{16\pi \varepsilon_o}\)

3. Two equal and opposite charges Q1 = 2 C and Q2 = -2C are placed at a distance of 6m from each other. What is the potential at the midpoint between the two charges?
a) 2 V
b) 0 V
c) 1 V
d) 3 V
View Answer

Answer: b
Explanation: Electric potential is a scalar quantity. So, the total potential will be the sum of all the individual charges.
V1 = \(\frac {kQ}{r} = \frac {2k}{3}\)
V2 = \(\frac {kQ}{r} = \frac {-2k}{3}\)
Total potential (V) = V1 + V2
V = k [ \((\frac {2}{3}) + (\frac {-2}{3})\) ]
V = 0 V
Therefore, the potential at the midpoint between the two charges is zero volt.
advertisement
advertisement

4. Electric field and electric field intensity are related to electric potential.
a) True
b) False
View Answer

Answer: a
Explanation: Yes, both are related to electric potential. The relationship between electric potential and electric field is a differential → electric field (E) is the gradient of electric potential (V) in the x direction. Thus, as the charge moves in the x direction, the rate of its change in potential is the value of the electric field. Then, electric field can be defined as the negative of the rate of derivative of potential difference.

5. The 3 charges → \(\frac {q}{2}\), -q, \(\frac {q}{2}\) are placed along the x axis at x = 0, x = r, x = 2 respectively. Find the resultant potential at a point A located at a distance y from charge q such that r << y.
a) V=\(\frac {4qr^2}{4\pi \varepsilon_o y}\)
b) V=\(\frac {qr^3}{4\pi \varepsilon_o y^3}\)
c) V=\(\frac {qr^2}{4\pi \varepsilon_o y^3}\)
d) V=\(\frac {qr^3}{4\pi \varepsilon_o y^2}\)
View Answer

Answer: c
Explanation: Electric potential at point A can be given by:
V=\(\frac {1}{4\pi \varepsilon_o}[(\frac {(\frac {q}{2})}{y+r}) – \frac {q}{y} + \frac {(\frac {q}{2})}{(y-r)}]\)
V=\(\frac {(qr^2)}{4\pi \varepsilon_o y(y^2 – r^2)}\)
V=\(\frac {qr^2}{4\pi \varepsilon_o y^3}\) (Since y >> r)
advertisement

Sanfoundry Global Education & Learning Series – Physics – Class 12.

To practice Physics Question Papers for Schools, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter