Question

In a uniform electric field, the potential is 10 V at the origin of coordinates, and 8V at each of the points (1, 0, 0), (0, 1, 0) and(0, 0, 1). The potential at the point (1, 1, 1) will be

Answer 1

Dgdgjd
દુઆઈહે gdgrbrud dybdv

Answer 2

Answer:4 V

Explanation:Let the electric field vector E→=Exi^+Eyj^+Ezk^E→=Exi^+Eyj^+Ezk^

VA−VB=−∫ABE→.drVA−VB=−∫BAE→.dr

V(1,0,0)−V(0,0,0)=−∫(1,0,0)(0,0,0)Ex=[−Exx](1,0,0)(0,0,0)V(1,0,0)−V(0,0,0)=−∫(0,0,0)(1,0,0)Ex=[−Exx](0,0,0)(1,0,0)

8−10=−Ex8−10=−Ex

Ex=2Ex=2

Similarly V(0,1,0)−V(0,0,0)=−∫(0,1,0)(0,0,0)EydyV(0,1,0)−V(0,0,0)=−∫(0,0,0)(0,1,0)Eydy

−2=−Ey×1−2=−Ey×1

Similarly Ez=2Ez=2

Thus

E→=2i^+2j^+2k^E→=2i^+2j^+2k^

Now

V(1,1,1)−V(0,0,0)=−∫(1,1,1)(0,0,0)(2i^+2j^+2k^).(dxi^+dyj^+dzk^)V(1,1,1)−V(0,0,0)=−∫(0,0,0)(1,1,1)(2i^+2j^+2k^).(dxi^+dyj^+dzk^)

=−∫(1,1.1)(0,0,0)(2dx+2dy+2dz)=−∫(0,0,0)(1,1.1)(2dx+2dy+2dz)

=−(2×1+2×1+2×1)=−(2×1+2×1+2×1)

V(1,1,1)−10=−6V(1,1,1)−10=−6

V(1,1,1)=4V.