The electric potential at x=3 on the curve of $y = x^{2}$ is 10V and it is 20 V at a point x= 4 (on the same curve). What is the average potential gradient between these two points?

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Solution

The correct option is B
The average potential gradient between two points say A and B is defined as the change of potential b/w the two points divided by the distance between them.
Since the points are on curve $y = x^{2}$ . For x=3, we get y=9 and for x=4, we get y= 16. Now the two pints between which we have to calculate the potential gradient becomes (3, 9) and (4, 16). Distance between these two points can be calculated using the distance formula that you might have studied in geometry.
[I am not providing the formula I recommend you to search about it on your own in case you don’t know it] Distance in this case = $5 \sqrt{2}$
Now, average potential gradient = $\frac{20 - 10}{2 \sqrt{5}} = \frac{10}{2 \sqrt{5}} = \sqrt{5}$

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The electric potential at x=3 on the curve of y=x2 is 10V and it is 20 V at a point x= 4 (on the same curve). What is the average potential gradient between these two points?

The electric potential at x=3 on the curve of $y = x^{2}$ is 10V and it is 20 V at a point x= 4 (on the same curve). What is the average potential gradient between these two points?