A charge of 3 coulomb moving in a uniform electric field experiences a force of 3000 newton. The potential…

A charge of 3 coulomb moving in a uniform electric field experiences a force of 3000 newton. The potential difference between the two points situated in a field at a distance of $1 \mathrm{~cm}$ from each other will be :
  1. 100
  2. 5000
  3. 10
  4. 50

Solution

Force on a charge in an electric field is $F=q E$
But $E=\frac{V}{d} \quad \therefore \quad F=\frac{q V}{d}$
or $\quad V=\frac{F d}{q}=\frac{300 \mathrm{~N} \times 1 \times 10^{-2} \mathrm{~m}}{3 C}=10$ volt ,

Asked in: JEE Mains - Electrostatics - Test 4

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