Question

Assume that an electric field Ė = 30x^2 j exists in a
space. Then the potential difference (VA - VB)
between the point A(1 m, 0 m) and B(0 m, 1 m) is​

Answer 1

The potential difference between point A and point B is -10 Volts

Explanation:

Given Electric field

$\vec E=30x^2\hat j$

Point A and B are

$\vec x_A=\hat i$

$\vec x_B=\hat j$

$\Delta x=\vec x_B-\vec x_A$

$\implies \Delta x=\hat j-\hat i$

or, $\vec dx=\hat j-\hat i$

$dV=-\vec E.\vec dx$

$\int_{V_A}^{V_B} dV=-\int_A^B (30x^2\hat j).(\hat j-\hat i)dx$

$\implies V_B-V_A=-\int_1^030x^2dx$

$\implies V_B-V_A=-30\frac{x^3}{3}\Bigr|_1^0$

$\implies V_B-V_A=10\times 1-0$

$\implies V_B-V_A=10$ V

$\implies V_A-V_B=-10$ V

Hope this answer is helpful.

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