Question

What is the magnitude of electric field at the point (2.95i-2.25j+4.10k)?

Answer 1

we know that E=-dv/dx

So we calculate in each direction electric field separately,

E=-(dv/dx,dv/dy,dv/dz)

Ex=-dV/dx

=>Ex=-d(–4x + 5y + 15z)/dx

=>Ex=4

Ey=-dV/dy

=>Ey=-d(–4x + 5y + 15z)/dy

=>Ey=-5

Ez=-dV/dz

=>Ez=-d(–4x + 5y + 15z)/dz

=>Ez=-15

E=4i-5j-15k

|E|=√(16+25+225) = 16.31 V/m

Cheers!!

Regards,

Vikas (B. Tech. 4th year

Thapar University)