The exponent follows from two facts put together: the depletion width grows as the square root of the bias, and capacitance is inversely proportional to that width.

Step 1 — the depletion width. Solving Poisson's equation across an abrupt junction gives

\(W=\sqrt{\dfrac{2\varepsilon_{s}}{q}\left(\dfrac{1}{N_{A}}+\dfrac{1}{N_{D}}\right)\left(V_{bi}+V_{R}\right)}\ \propto\ V_{R}^{1/2}\)

Step 2 — the capacitance. The depletion region is a slab of insulating, carrier-free silicon between two conducting neutral regions — a parallel-plate capacitor whose dielectric thickness varies with bias:

\(C_{j}=\dfrac{\varepsilon_{s}A}{W}\ \propto\ \dfrac{1}{V_{R}^{1/2}}=V_{R}^{-1/2}\)

which is option 3.

Option 4 is the deliberate near-miss : the exponent \(-1/3\) is correct, but for a linearly graded junction, where the doping changes gradually across the metallurgical boundary. The question specifies "abrupt… with uniform doping on either side", which selects the square-root law. The general form is written

\(C_{j}=\dfrac{C_{j0}}{\left(1+\dfrac{V_{R}}{V_{bi}}\right)^{n}}\)

with \(n=1/2\) for abrupt and \(1/3\) for graded profiles.

Why the effect is useful. A reverse-biased junction is an electrically variable capacitor — the varactor. Placed in the tank of an oscillator it makes the frequency a function of a control voltage, which is the basis of every VCO, FM modulator and PLL frequency synthesiser. Hyperabrupt profiles are manufactured deliberately, with an exponent near 2, because that makes \(f=1/2\pi\sqrt{LC}\) vary almost linearly with the tuning voltage.

Where it is a nuisance : the same capacitance limits switching speed, since it must be charged and discharged at every transition — which is why fast diodes are made with small junction areas.

Hence, C_j ∝ V_R^−1/2.