Explanation:
It is given that,
Charge in the particle, [tex]q=6.4\ nC=6.4\times 10^{-9}\ C[/tex]
Work done by the force, [tex]W_1=6.5\times 10^{-5}\ J[/tex]
The kinetic energy of the particle, [tex]E_k=3.7\times 10^{-15}\ J[/tex]
(a) Using the conservation of energy to find the work done by the electric force. Let it is given by [tex]W_2[/tex] as :
[tex]W_1+W_2=E_k[/tex]
[tex]W_2=E_k-W_1[/tex]
[tex]W_2=3.7\times 10^{-15}-6.5\times 10^{-5}[/tex]
[tex]W_2=-6.49\times 10^{-5}\ J[/tex]
(b) Let V is the electric potential. the work done by the electric charge per unit positive charge is called its electric potential. Mathematically,
[tex]\Delta V=\dfrac{W_2}{q}[/tex]
[tex]\Delta V=\dfrac{-6.49\times 10^{-5}}{6.4\times 10^{-9}}[/tex]
[tex]\Delta V=10.14\times 10^3\ volts[/tex]
[tex]\Delta V=10.14\ kV[/tex]
Hence, this is the required solution.
