Question

Determine the electric field strength vector if the potential of the field depends on x,y coordinates as V=10axy

Answer 1

$\mathfrak{\underline{\underline{Solution:-}}}$

$\bf{Electric\;field\;strength\;vector:}$

$\sf{\vec{E}=-\frac{d}{dx}(10axy)\hat{i}-\frac{d}{dy}(10axy)\hat{j}-\frac{d}{dz}(10axy)\hat{k}}$

$\sf{\vec{E}= -10ay\;\hat{i}-10ax\;\hat{j}}$

★${\boxed{\boxed{\bf{\vec{E}= -\bigg[ (10ay)\;\hat{i}+(10ax)\;\hat{j}\bigg]}}}$

Answer 2

electric field strength in vector , E = -10a(y i + x j)

we know, electric field is the negative of rate of change of potential difference with respect to position.

i.e., E = -dV/dx

here potential of the field depends on x, y co-ordinates as V = 10axy

so we have to apply partial differentiation concept.

i.e., E = - δv/δx i - δy/δy j - δv/δz k

= - 10ay i - 10a x j

= -10a (y i + x j)

magnitude of electric field, |E| = 10a√(x² + y²)

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