Answer:
Charge, [tex]q=5.37\times 10^{-19}\ C[/tex]
Explanation:
Given that,
Distance moved by particle, d = 12 m
Electric field strength, E = 76 N/C
Decrease in potential energy, [tex]P=4.9\times 10^{-16}\ J[/tex]
We need to fine the charge on the particle. The change in potential energy of the system is given by :
[tex]P=qEd[/tex]
q is charge
[tex]q=\dfrac{P}{Ed}\\\\q=\dfrac{4.9\times 10^{-16}}{76\times 12}\\\\q=5.37\times 10^{-19}\ C[/tex]
So, the charge on the particle is [tex]5.37\times 10^{-19}\ C[/tex].
