Question

If potential function in an electric field is given as v= -4x^2+ 3y^2+2z , then electric field at point (1,0,2)

Answer 1

E=-dV/dr

Find derivative with respect to each function and substitute

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Answer 2

Given:

• The potential in an electric field, V = $-4x^2+3y^2+2z$

To Find:

• The electric field at points (1,0,2)

Solution:

The formula to find the electric field is given by,

E = -dV/dr = -($\frac{dV_x}{dx} +\frac{dV_y}{dy} +\frac{dV_z}{dz}$) → {equation 1}

On substituting the given values in equation 1 we get,

⇒ E = -($\frac{-4x^2}{dx} +\frac{3y^2}{dy} +\frac{2z}{dz}$)

On partially differentiating the above formula respectively with x,y, and z we get,

⇒ E = -(-8x+6y+2)

Now substitute (x,y,z) = (1,0,2) in the above formula we get,

⇒ E = -(-8(1)+6(0)+2)

⇒ E = -(-8+2) = 8-2 {multiply the subtract term inside and subtract the terms}

⇒ E = 6

∴ The value of electric field at point (1,0,2) = 6.