JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Electrostatics

Both forces fall off as $1/r^2$, so the separation $r$ cancels in the ratio and its value is not needed:

$\frac{F_e}{F_g}=\frac{\frac{1}{4\pi {\epsilon }_0}\frac{q_A{q}_B}{r^2}}{\frac{G{m}_A{m}_B}{r^2}}=\frac{k{q}_A{q}_B}{G{m}_A{m}_B}$

Substituting:

$\frac{F_e}{F_g}=\frac{\left(9\times 10^9\right)\left(6.67\times 10^{-19}\right)\left(9.6\times 10^{-10}\right)}{\left(6.67\times 10^{-11}\right)\left(19.2\times 10^{-27}\right)\left(9\times 10^{-27}\right)}$

Group the numerical coefficients and the powers of ten separately:

coefficients: $\frac{9\times 6.67\times 9.6}{6.67\times 19.2\times 9}=\frac{9.6}{19.2}=\frac{1}{2}$

(the 9 and the 6.67 cancel exactly — the data is chosen so that $q_A$ cancels against $G$)

powers of ten: $10^{9-19-10}÷10^{-11-27-27}=10^{-20}÷10^{-65}=10^{45}$

$\therefore \frac{F_e}{F_g}=\frac{1}{2}\times 10^{45}=5\times 10^{44}$

Comparing with $P\times 10^{44}$:

$P=5$

Note: the electrostatic force here exceeds the gravitational force by about 44 orders of magnitude — the standard illustration of why gravity is irrelevant at the atomic scale and why matter must be very nearly electrically neutral in bulk.