Question

A uniform electric field of 400 V/m is directed at 45 degree angle from positive x-axis as shown in figure. the potential difference Va-vb is given by

Answer 1

Answer:

Explanation:

=> E = Ex(i)+Ey(j)

E=Ecos45⁰(i)+Esin45⁰(j)

=E/√2(i) + E/√2(j)

= E/√2(i+j)

V_A-V_B = $\int\limits^B_A {E} di =  \int\limits^B_A {E/\sqrt{2} (i+j) (dn i + dn j)}$

=$E/\sqrt{2} [ \int\limits^{0.03}_0 dx + \int\limits^{0}_{0.02} dy ]\\= E/\sqrt{2} [0.03-0.02]$

= 400/√2 * 0.01

= 400 * 0.01/1.414

=2.83 V

Thus, the potential difference Va-vb is given by 2.83 V