A parallel plate capacitor consists of two plates of 2m *1m. The space...
Potential Gradient of a Parallel Plate Capacitor
Given:Area of each plate, A = 2m * 1m = 2m
2Distance between plates, d = 1mm = 0.001m
Relative permittivity of dielectric, ε
r = 7
Potential difference, V = 300V
Formula:The potential gradient is given by the formula:
E = V/d
where E is the electric field strength and d is the distance between the plates.
Solution:Substituting the given values in the formula:
E = V/d = 300V/0.001m = 300000V/m
Since the electric field strength is the negative of the potential gradient, the potential gradient is:
|E| = |dV/dx| = 300000V/m
Therefore, the potential gradient of the parallel plate capacitor is 300000V/m.
Explanation:A parallel plate capacitor consists of two parallel plates with equal and opposite charges. When a potential difference is applied across the plates, an electric field is created between the plates. The strength of the electric field is proportional to the potential difference and inversely proportional to the distance between the plates.
The potential gradient is the rate at which the potential changes with respect to distance. It is the magnitude of the electric field strength. In a parallel plate capacitor, the potential gradient is constant between the plates and is given by the formula E = V/d, where V is the potential difference and d is the distance between the plates.
In this problem, the potential gradient is calculated using the given values of the area of the plates, the distance between the plates, the relative permittivity of the dielectric, and the potential difference. The potential gradient is found to be 300000V/m, which means that the potential difference changes by 300000V for every meter of distance between the plates.