Question

The potiential V due to a charge distribution at point (x,y) is given by V= -4x^2+3y. Calculate electric field in magnitude and direction due to charge distribution at the point (1,1)
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Answer 1

Answer:

$E= 8.653\,N/C\\\\\theta = 20.55^{o}~from~+ve ~~x-axis~in~IV^{th}~quadrant$

Explanation:

Electric point at any point is given by

$E = \frac{-dV}{dr}$

Hence, here, in x-direction,

$E_{x}=\frac{-d(-4x^{2}+3y)}{dx}\\\\E_{x}=8x$

And, in y-direction,

$E_{y}=\frac{-d(-4x^{2}+3y)}{dy}\\\\E_{y}=-3$

Hence, at point (1,1)

$E_{x} = 8(i)\,N/C\\E_{y}=-3(j)\,N/C$

Direction of it's resultant will be

tanθ = 3/8          (θ is in $IV^{th}$ quadrant from +ve x-axis)

θ = 20.55°

So, it's magnitude will be vector sum of these components i.e.,

$E=\sqrt{E_{x}^{2}+E_{y}^{2}+2cos\theta}\\\\E=\sqrt{8^{2}+(-3)^{2}+2*0.936}\\\\E=\sqrt{64+9+1.872}\\\\E=\sqrt{74.872}\\\\E= 8.653\,N/C$