PHYSICS
JAMB 2003 - Question 33
Physics 2003 JAMB Past Questions - Question 33: An electron of charge 1.6 x 10¹ is accelerated between two metal plates. If the kinetic energy of the electron is 4.8 x 10¹J, the potential difference between the plates is
Choose the correct answers from the options given.
A:
B:
C:
D:
Correct Answer
C
Explanation
To find the potential difference between the plates, you can use the formula for the kinetic energy of a charged particle in an electric field:
K.E. = q * V
Where:
K.E. is the kinetic energy of the electron (4.8 x 10^16 J)
q is the charge of the electron (1.6 x 10^-19 C)
V is the potential difference between the plates (which is what we want to find)
Now, you can rearrange the formula to solve for V:
V = K.E. / q
Plug in the values:
V = (4.8 x 10^16 J) / (1.6 x 10^-19 C)
Now, perform the division:
V = (4.8 x 10^16 J) / (1.6 x 10^-19 C) = 3 x 10^35 V
So, the potential difference between the plates is approximately 3 x 10^35 volts.
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