CBSE Previous Year Questions Electric Charges and Fields - Physics Class 12 - NEET

Multiple Choice Type Questions

Q1: Two charges –q each are placed at the vertices A and B of an equilateral triangle ABC. If M is the mid-point of AB, the net electric field at C will point along      [1 Mark]
(A) CA
(B) CB
(C) MC
(D) CM
Ans: (D) CM
Sol: At point C, electric field due to each −q charge is directed towards the charge.

• Field due to charge at A → along CA
• Field due to charge at B → along CB

Since triangle ABC is equilateral, these two fields have equal magnitude and are symmetrically inclined.
Their horizontal components cancel, and vertical components add along CM.

Q2: Figure shows variation of Coulomb force (F) acting between two point charges with \(\frac{1}{r^2}\), r being the separation between the two charges (q₁, q₂) and (q₂, q₃). If q₂ is positive and least in magnitude, then the magnitudes of q₁, q₂ and q₃ are such that   [1 Mark]

(a) \(q_2 < q_3 < q_1\)
(b) \(q_3 < q_1 < q_2\)
(c) \(q_1 < q_2 < q_3\)
(d) \(q_2 < q_1 < q_3\)
Ans: (d) \(q_2 < q_1 < q_3\)
Sol:  
$F = k \dfrac{q_i q_j}{r^2}$
The graph of F vs 1/r² has slope = kq_iqj.

• Line (q₁, q₂) has positive slope ⇒ q₁ and q₂ have same sign.
• Line (q₂, q₃) has negative slope ⇒ q₂ and q₃ have opposite signs.
• Given q₂ is positive and least in magnitude.
• Steeper line for (q₁, q₂) ⇒∣q₁q₂∣ > ∣q₂q₃∣ ⇒ $|q₁|\; >\; |q₃|$∣q₁∣ > ∣q₃∣.

Since q₂ is least: q₂ < q₁ < q₃

Q3: Which one of the following statements is correct?    [1 Mark]
Electric field due to static charges is
(a) conservative and field lines do not form closed loops.
(b) conservative and field lines form closed loops.
(c) non-conservative and field lines do not form closed loops.
(d) non-conservative and field lines form closed loops.
Ans: (a) conservative and field lines do not form closed loops.
Sol: Static electric fields are conservative, and field lines start from positive charges and end on negative charges (or go to infinity). They do not form closed loops (closed loops are characteristic of induced electric fields).

Q4: Two identical point charges are placed at the two vertices A and B of an equilateral triangle of side \(l\). The magnitude of the electric field at the third vertex P is E. If a hollow conducting sphere of radius \((l / 4)\) is placed at P, the magnitude of the electric field at point P now becomes    [1 Mark]
(a) Greater than E
(b) E
(c) \(\frac{E}{2}\)
(d) zero
Ans: (d) Zero
Sol: Inside a hollow conducting sphere, the electric field is zero.

Q5: Consider two identical dipoles D₁ and D₂ . Charges -q and q of dipole D₁ are located at (0, 0) and (a, 0) and that of dipole D₂ at (0, a) and (0, 2a) -L Z in x-y plane, respectively. The net dipole moment of the system is    [1 Mark]
(a) \(qa(\hat{i} + \hat{j})\)
(b) \(qa(\hat{i} - \hat{j})\)
(c) -\(qa(\hat{i} + \hat{j})\)
(d) -\(qa(\hat{i} - \hat{j})\)
Ans: (a) \(qa(\hat{i} + \hat{j})\)
Sol: Dipole moment $\vec{p} = q \vec{d}$ ; (from $-q$ to $+q$).

• item For $D_1$: charges at $(0,0)$ and $(a,0)$ $\rightarrow$ along $+x$ direction
      \(\vec{p}_1 = qa\,\hat{i}\)
• item For $D_2$: charges at $(0,a)$ and $(0,2a)$ $\rightarrow$ along $+y$ direction
      \(\vec{p}_2 = qa\,\hat{j}\)

Net dipole moment:
\(\vec{p} = \vec{p}_1 + \vec{p}_2 = qa(\hat{i} + \hat{j})\)

Q6: A body acquires charge \(8.0 \times 10^{-12} \, \mathrm{C}\). The mass of the body:   [1 Mark]
(a) increases by \(4.5 \times 10^{-7} \, \mathrm{kg}\)
(b) decreases by \(1.0 \times 10^{-6} \, \mathrm{kg}\)
(c) decreases by \(4.55 \times 10^{-23} \, \mathrm{kg}\)
(d) increases by \(9.1 \times 10^{-23} \, \mathrm{kg}\)
Ans: (c)
Sol: Positive charge means loss of electrons. Mass decreases by the mass of electrons lost.

• Number of electrons lost = \(\frac{Q}{e} = \frac{8.0 \times 10^{-12}}{1.6 \times 10^{-19}} = 5.0 \times 10^{7}\)
• Mass decrease = \((5.0 \times 10^{7}) \times (9.1 \times 10^{-31}) = 4.55 \times 10^{-23} \, \mathrm{kg}\).

Q7: Two point charges \(Q\) and \(-q\) are held 'r' distance apart in free space. A uniform electric field \(\overrightarrow{E}\) is applied in the region perpendicular to the line joining the two charges. Which one of the following angles will the direction of the net force acting on charge \(-q\) make with the line joining \(Q\) and \(-q\)?    [1 Mark]
\((a) \quad \tan^{-1}\frac{4\pi\epsilon_0 E r^2}{Q}\)
\((b) \quad \cot^{-1}\frac{4\pi\epsilon_0 E r^2}{Q}\)
\((c) \quad \tan^{-1}\frac{QE}{4\pi\epsilon_0 r^2}\)
\((d) \quad \cot^{-1}\frac{QE}{4\pi\epsilon_0 r^2}\)
Ans: (a) \(\tan^{-1}\frac{4\pi\epsilon_0 E r^2}{Q}\)
Sol: 
On charge −q: