The electric potential at a point in free space due to a charge $Q$ coulomb is $Q imes 10^{11}$ V. The elect

Question

The electric potential at a point in free space due to a charge $Q$ coulomb is $Q 	imes 10^{11}$ V. The electric field at that point is:

Accepted Answer

1.  **Formulas:**
    *   Electric potential $V$ due to a point charge $Q$ at distance $r$: $V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$ [NCERT Class 12, Sec 2.3, Eq 2.8].
    *   Electric field magnitude $E$ due to a point charge $Q$ at distance $r$: $E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2}$ [NCERT Class 12, Sec 1.8, Eq 1.6].
    *   Relationship: $E = \frac{V}{r}$.

2.  **Calculation:**
    *   Given $V = Q 	imes 10^{11}$ Volts.
    *   Substitute into the potential formula to find $r$:
        $Q 	imes 10^{11} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$
        $r = \frac{1}{4\pi\epsilon_0 	imes 10^{11}}$
    *   Substitute $r$ into the field relationship $E = V/r$:
        $E = \frac{Q 	imes 10^{11}}{\frac{1}{4\pi\epsilon_0 	imes 10^{11}}}$
        $E = (Q 	imes 10^{11}) 	imes (4\pi\epsilon_0 	imes 10^{11})$
        $E = 4\pi\epsilon_0 Q 	imes 10^{22} 	ext{ V/m}$

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