The variation of electric potential with distance d from a fixed point is as shown in the figure. The electric intensity at d = 5m is
Question
The variation of electric potential with distance d from a fixed point is as shown in the figure.
The electric intensity at d = 5m is:
Solution
The electric field intensity (E) is related to the electric potential (V) by the relation:
E = -dV/dr
where dV/dr is the rate of change of potential with distance.
From the graph, we can see that the potential decreases by 100V as the distance increases from 3m to 7m. Therefore, the rate of change of potential with distance is:
dV/dr = (V2 - V1) / (r2 - r1) = (0 - 100) / (7 - 3) = -25 V/m
Therefore, the electric field intensity at d = 5m is:
E = -dV/dr = -(-25) = 25 N/C
So, the electric field intensity at d = 5m is 25 N/C.
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