The distance between plates of a parallel plate capacitor is 5d. The positively charged plate is at x = 0 and negatively charged plates is at x = 5d. Two slabs – one of conductor and the other of a dielectric, both of same thickness d are inserted between the plates as shown in figure. Potential (V) versus distance x graph will be:
Question
The distance between plates of a parallel plate capacitor is 5d. The positively charged plate is at x = 0 and negatively charged plates is at x = 5d. Two slabs – one of conductor and the other of a dielectric, both of same thickness d are inserted between the plates as shown in figure. Potential (V) versus distance x graph will be:
Solution
The potential V versus distance x graph for this setup will be a piecewise linear function. Here's how it will look:
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From x = 0 to x = d (the conductor slab): The potential will be constant because the electric field inside a conductor in electrostatic equilibrium is zero. So, the potential does not change across the conductor. If we assume the potential at x = 0 to be V0, then the potential from x = 0 to x = d will be V0.
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From x = d to x = 2d (the dielectric slab): The potential will decrease linearly. This is because the electric field in a dielectric is constant and directed from the positive to the negative plate. Therefore, the potential decreases linearly across the dielectric. Let's say the potential decreases to V1 at x = 2d.
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From x = 2d to x = 5d (the vacuum): The potential will continue to decrease linearly. The electric field in a vacuum is also constant and directed from the positive to the negative plate. Therefore, the potential continues to decrease linearly across the vacuum. The potential will reach zero at x = 5d.
So, the graph of V versus x will be a horizontal line from x = 0 to x = d, a straight line decreasing from x = d to x = 2d, and another straight line decreasing from x = 2d to x = 5d.
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