Question

A particle with a charge of 6.40 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved a distance of 5.00 cm , the additional force has done an amount of work equal to 6.50×10−5 J and the particle has kinetic energy equal to 3.70×10−5 J.
(a) What work was done by the electric force?
(b) What is the potential of the starting point with respect to the end point?

Answer 1

Explanation:

It is given that,

Charge in the particle, $q=6.4\ nC=6.4\times 10^{-9}\ C$

Work done by the force, $W_1=6.5\times 10^{-5}\ J$

The kinetic energy of the particle, $E_k=3.7\times 10^{-15}\ J$

(a) Using the conservation of energy to find the work done by the electric force. Let it is given by $W_2$ as :

$W_1+W_2=E_k$

$W_2=E_k-W_1$

$W_2=3.7\times 10^{-15}-6.5\times 10^{-5}$

$W_2=-6.49\times 10^{-5}\ J$

(b) Let V is the electric potential. the work done by the electric charge per unit positive charge is called its electric potential. Mathematically,

$\Delta V=\dfrac{W_2}{q}$

$\Delta V=\dfrac{-6.49\times 10^{-5}}{6.4\times 10^{-9}}$

$\Delta V=10.14\times 10^3\ volts$

$\Delta V=10.14\ kV$

Hence, this is the required solution.