CBSE Class 12 Science Physics · Ch 1: Electric Charges and Fields

Topics in this test

Difficulty: 3 easy · 9 medium · 18 hard

Sample questions with answers

The first 5 questions of this free test.

1. Q1. Two identical conducting spheres A and B carry charges $+6Q$ and $-2Q$ respectively. A third identical uncharged sphere C is first touched to A, then to B, and finally removed. What is the final charge on sphere B?

   • A. $+\frac{3Q}{2}$
   • B. $-\frac{Q}{2}$
   • C. $+2Q$
   • D. $+\frac{Q}{2}$
   Solution

   Touching C to A splits $+6Q$ equally, giving each $+3Q$; then C ($+3Q$) touches B ($-2Q$), so their total $+Q$ splits equally, giving B $+\frac{Q}{2}$.

2. Q2. Two point charges $+Q$ and $+Q$ are separated by a distance $d$. A third charge $q$ is placed on the line joining them so that the system is in equilibrium. The position and nature of $q$ are:

   • A. at the midpoint and negative, of magnitude $Q/2$
   • B. at the midpoint and positive, of magnitude $Q/4$
   • C. at the midpoint and negative, of magnitude $Q/4$
   • D. at one-third distance from $+Q$ and positive, of magnitude $Q$
   Solution

   By symmetry $q$ must sit at the midpoint, and for the two positive charges themselves to be in equilibrium the middle charge must be negative; its magnitude works out to $Q/4$.

3. Q3. An infinitely long straight wire has a uniform linear charge density $\lambda$. If the perpendicular distance from the wire is doubled, the magnitude of the electric field becomes:

   • A. one-fourth of its original value
   • B. double its original value
   • C. unchanged
   • D. half of its original value
   Solution

   For a line charge $E=\frac{\lambda}{2\pi\varepsilon_0 r}$, which is inversely proportional to $r$; doubling $r$ halves the field.

4. Q4. Assertion (A): The electric field inside a uniformly charged thin spherical shell is zero everywhere. Reason (R): The net charge enclosed by any Gaussian surface drawn inside the shell is zero. Choose the correct option.

   • A. A is true but R is false
   • B. A is false but R is true
   • C. Both A and R are true and R is the correct explanation of A
   • D. Both A and R are true but R is not the correct explanation of A
   Solution

   A Gaussian surface inside the shell encloses no charge, so by Gauss's law the flux and hence the field is zero; both statements are true and R correctly explains A.

5. Q5. The electric field due to a short dipole at a point on its axial line at distance $r$ is $E_1$, and at the same distance on its equatorial line is $E_2$. The ratio $E_1:E_2$ is:

   • A. $1:1$
   • B. $4:1$
   • C. $2:1$
   • D. $1:2$
   Solution

   Axial field is $\frac{2kp}{r^3}$ and equatorial field is $\frac{kp}{r^3}$, so their ratio is $2:1$.

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