Electrostatics MCQs for NEET — Physics Questions with Answers

The potential at a point (0, 0, z) due to a dipole formed by charges -q at (0, 0, -a) and +q at (0, 0, a) can be calculated using the formula for the potential due to a dipole. The correct formula for the potential at a distance z from the dipole center along the axis is V = (1/4πε₀) * (2qa / (z² - a²)). Therefore, the correct answer is $\frac{2qa}{4\pi\varepsilon_0(z^2 - a^2)}$.

Potential will be zero at two points $ At internal point (M ) = {1 \over 4 \pi \varepsilon} \left[ { 2 \times 10^{-6} \over (6 - l ) }+ { ( -1 \times 10^{-6} )\over l } \right] = 0 \Rightarrow l = 2 $ So distance of M from origin; x = 6 -2 = 4 At exterior point (N ) $ {1 \over 4 \pi \varepsilon} \left[ { 2 \times 10^{-6} \over (6 - l ) }+ { ( -1 \times 10^{-6} \over l } \right] = 0 \Rightarrow l' = 6 $

To find the number of identical mercury drops that combine to form one large drop, we use the relationship between the potential and the radius. When N drops combine, the total volume remains the same, so if each drop has radius r, the large drop will have radius $R = rN^{1/3}$. Potential V is proportional to the charge and inversely proportional to the radius. Given initial potential V = 10V and final potential 40V, we set up the equation $10N^{1/3} = 40$, which simplifies to $N^{1/3} = 4$, giving N = 64. Therefore, the closest option is 8, making option o3 the correct answer.

When two spherical conductors are brought into contact, charge is redistributed until they reach the same potential. The common potential $V$ is given by the formula: $$ V = rac{r_1 V_1 + r_2 V_2}{r_1 + r_2} $$ Here, $r_1$ and $r_2$ are the radii of the conductors, and $V_1$ and $V_2$ are their respective potentials.

The capacitance $C$ of a parallel plate capacitor with a dielectric slab of thickness $t$ and dielectric constant $K$ inserted can be calculated using the formula: $$ C' = rac{ ext{original capacitance} imes ext{dielectric constant}}{1 + rac{t}{d}(K - 1)} $$ Given that $C' = rac{4}{3}C$ and $t = rac{d}{2}$, solving this equation yields a dielectric constant $K$ of 2.5.

To maintain the same potential difference, the effective distance $d_{eff}$ between the plates of the capacitor is given by: $$ d_{eff} = d - t + rac{t}{K} $$ Given that $d - t + rac{t}{K} = d + 1.6 ext{ mm}$ and $t = 2 ext{ mm}$, solving this equation yields a dielectric constant $K$ of 5.

When capacitors are connected in series, the equivalent capacitance \( C_s \) is given by: $$ rac{1}{C_s} = rac{1}{4} + rac{1}{6} + rac{1}{12} = rac{1}{2} \rightarrow C_s = 2 ext{ F} $$ When capacitors are connected in parallel, the equivalent capacitance \( C_p \) is given by: $$ C_p = 4 + 6 + 12 = 22 ext{ F} $$ The ratio of the equivalent capacitance in series to that in parallel is: $$ rac{C_s}{C_p} = rac{2}{22} = rac{1}{11} ). Therefore, the ratio is 1:11.