A point charge \(Q\) is at distance \(r\) from a point P. Match the quantity in List-I with the change in it in List-II.
| List-I | List-II |
|---|---|
| (A) Potential at P when \(Q\) is doubled and \(r\) is halved | (I) Becomes half |
| (B) Magnitude of the electric field at P when \(Q\) is doubled and \(r\) is doubled | (II) Becomes four times |
| (C) Magnitude of the electric field at P when \(Q\) is halved and \(r\) is halved | (III) Remains the same |
| (D) Potential at P when both \(Q\) and \(r\) are doubled | (IV) Becomes double |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
For a point charge, the potential is \( V=Q/4\pi\varepsilon_0r \) and the field is \( E=Q/4\pi\varepsilon_0r^{2} \). So \( V \) varies as \( Q/r \) and \( E \) varies as \( Q/r^{2} \).
(A) Potential at P when \( Q \) is doubled and \( r \) is halved — This matches (II) becomes four times. The potential now varies as \( 2Q/(r/2)=4Q/r \).
(B) Magnitude of the electric field at P when \( Q \) is doubled and \( r \) is doubled — This matches (I) becomes half. The field now varies as \( 2Q/(2r)^{2}=Q/2r^{2} \).
(C) Magnitude of the electric field at P when \( Q \) is halved and \( r \) is halved — This matches (IV) becomes double. The field now varies as \( (Q/2)/(r/2)^{2}=2Q/r^{2} \).
(D) Potential at P when both \( Q \) and \( r \) are doubled — This matches (III) remains the same. The ratio \( Q/r \) does not change.
NCERT (2026-27):
Summary – “… If potential at infinity is chosen to be zero; potential at a point with position vector r due to a point charge Q placed at the origin is given is given by”
\( \displaystyle V(r)=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r} \)
Potential Due to a Point Charge – “… shows how the electrostatic potential ( ∝ 1/r) and the electrostatic field ( ∝ 1/r2 ) varies with r.”