Q.In the following, if the distance of observation point from the centre of source becomes half, then how much electric field be changed and why?

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

…

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 $r' = 2r$.

…

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 $10^{42}$.

…

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): $\dfrac{F_e}{F_g}=\dfrac{e^2/4\pi\varepsilon_0}{Gm_em_p}\approx 2.3\times10^{39}$

…

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,

…