The magnitude of the electric field between two charged, parallel plates is 3.65E7 N/C. If the potential energy of 2.35uC charges, which is located between the plates is 0.25J what is the separation of the plates?
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Answer:Distance of separation between the plates of a parallel plate capacitor d= 0.005m
Explanation:Given
Electric field E =3.65E7 N/C
Potential energy U =0.25J
Charge Q =2.35uC
To find the distance of separation d
We know that
U = QEd /2
d = 2U /QE
= 2 *0.25 /(3.65E7 *2.35u)
= 0.005m is the distance of separation between the two plates of a parallel plate capacitor.
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Given :
Electric field (E) = 3.6537 N/C
Charge (q) = 2.35μC
Potential energy (U) = 0.25J
To find :
The separation between the plates = ?
Solution :
- Potential energy U = qEd/2
d = 2U/qE
- By substituting the values
d = (2×0.25) / (2.35×10⁻⁶×3.6537)
d = 0.5 / 8.556195
d = 0.0584×10⁶ m
The separation between the plates is 0.0584×10⁶ m
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