Q.Two charged metallic spheres with radii $R_1$ and $R_2$ respectively are brought in contact and then separated they carry same charge after separation. What is the ratio of electric fields at the surface of two sphere after separation?

By Coulomb's law the field of a point charge is $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$, so $E\propto 1/r^2$.

The force on a small test charge $q_0$ placed at distance $r$ from a point charge $q$ is $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{qq_0}{r^2}$ (Coulomb's law). The electric field is defined as force per unit test charge, $E=F/q_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$. Since $q$ and the constant $\dfrac{1}{4\pi\varepsilon_0}$ do not depend on $r$, the field varies only through the $1/r^2$ te …

Coulomb's force follows an inverse-square law, so doubling the separation cuts the force to one-quarter.

Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2} = \dfrac{k q_1 q_2}{r^2}$

At separation $r$: $F = \dfrac{k q_1 q_2}{r^2}$

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By Coulomb's law $F \propto 1/r^2$; halving the separation $r$ quadruples the force.

Coulomb's law states

$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$

If the distance is halved, $r' = r/2$, then …

Surface field $E=kQ/R^2$; with equal charge $Q$ on each sphere, $E \propto 1/R^2$.

The electric field at the surface of a charged conducting sphere of radius $R$ carrying charge $Q$ is $E = \dfrac{kQ}{R^2}$. Given that after separation both spheres carry the same charge $Q$ (as stated), the field at each surface is inversely proportiona …

Doubling both charges and the distance leaves the electrostatic force unchanged because the charge-product factor of 4 exactly cancels the distance-squared factor of 4.

Coulomb's law: $F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = k\frac{q_1q_2}{r^2}$

Let the new charges be $q_1' = 2q_1$, $q_2' = 2q_2$, and the new separation $r' = 2r$.

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The electric field of a point charge falls off as the inverse square of distance.

By Coulomb's law, the electric field due to a point charge $q$ at a distance $r$ is

$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ …

For two electrons, $F_e/F_g \approx 4\times10^{42}$ — order 10⁴².

Both forces between two identical charges (electrons) obey an inverse-square law, so the distance cancels in the ratio:

$\frac{F_e}{F_g} = \frac{\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}}{G\dfrac{m_e^2}{r^2}} = \frac{e^2}{4\pi\varepsilon_0 G m_e^2}.$

Putting $e = 1.6\times10^{-19}$ C and $m_e = 9.1\times10^{-31}$ kg gives $\approx 4\times10^{42}$ — i.e. of order $10^{42}$.

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Coulomb's law force between an electron and proton vastly exceeds Newton's gravitational force between them, because e²/(4πε₀) is enormously larger than Gm_em_p.

Electrostatic force: $F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}$. Gravitational force: $F_g=\dfrac{Gm_em_p}{r^2}$.

Taking the ratio (the $r^2$ cancels): $\dfrac{F_e}{F_g}=\dfrac{e^2/4\pi\varepsilon_0}{Gm_em_p}\approx 2.3\times10^{39}$

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The forces are an action–reaction pair, so they are equal in magnitude: ratio 1 : 1, option (C).

The Coulomb force each charge exerts on the other is

$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}.$

Both charges share the same product $q_1 q_2$ and the same separation $r$, so the magnitude is identical for each. This is also required by Newton's third law: the force on charge 1 due to charge 2 is equal and …

By Coulomb's law $F\propto 1/r^2$; reducing distance to one-fourth increases force 16 times.

Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$, so $F \propto \dfrac{1}{r^2}$.

If new distance $r' = r/4$: …

By Coulomb's law, force varies as $1/r^2$, so doubling $r$ divides the force by 4.

Coulomb's law gives the force between two point charges $q_1, q_2$ separated by distance $r$ as

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r^2}$

If the separation is doubled, $r \to 2r$, then …

By Coulomb's law the field of a point charge is E = kQ/r², so E ∝ 1/r².

The electric field intensity at a point is the force per unit positive test charge. For a point charge Q, the magnitude of the field at distance r is

E = (1/4πε₀) × (Q / r²) = kQ/r²,

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