Question

if potential at a point p(1 1 1) is given by the relation V=x^2-y^2+2z, then calculate the electric field at p.

Answer 1

As shown in above image, Electric field is given by,

$E \: = \frac{dV}{dx} + \frac{dV}{dy} \: + \frac{dV}{dz}$

So, now differentiating with respect x, y & z. Then their summation will give equation of Electric field.

$E = \frac{d( {x}^{2} - {y}^{2} + 2z)}{dx} + \frac{d({x}^{2} - {y}^{2} + 2z)}{dy} + \frac{({x}^{2} - {y}^{2} + 2z)}{dz}$

$E \: = 2x \: - 2y \: + 2$
Now for electric field at (1,1,1)

$E \: = 2(1) - 2(1) + 2$

$E \: = 2 \: \frac{N}{C}$

$E \: = -2 \: \frac{N}{C}$

Note:- Here minus sign shows the direction.