Question

two point charges A and B are 2 cm apart and a uniform electric field acts along the straight line AB directed from A to B with E= 200 N/C. a particle of charge +10(-6)C is taken from A to B along AB. calculate the force on the charge, the potential difference and work done on the charge by E.

Answer 1

Given :

Distance between the charges = 2 cm

Magnitude of the electric field = 200 N/C

Charge of the particle = 10⁻⁶ C

To Find :

The force on the charge

The potential difference

The work done on the charge by E

Solution :

• The electrostatic force on the charged particle is given by the formula

           $F=qE\\ F=10^{-6}\times200\\ F=2\times10^{-4}N$

  The force on the particle is 2×10⁻⁴ N.

• In a uniform electric field the potential difference between two points is given by

           $V = Ed\\ V=200\times2\times10^{-2}\\ V=4V$

 The potential difference between the points is 4V.

• The work done on an object is defined as the product of the force and the distance moved by the object

           $W=F\times d\\ W=2\times10^{-4}\times2\times10^{-2}\\ W=4\times10^{-6}J$

 The work done on the charged particle is 4×10⁻⁶J.