Question

Electric field in a region of space is given by
ILLUMIINNIT
E = 2xi + 3y(V/m). Magnitude of potential
difference between the points (1, 2) & (3,
WA
is a volt. Fill
Alpha/10​

Answer 1

Explanation:

We know that potential difference V is related to electric field E by the relation

E=-dV/dr

or dV=-Edr

So potential difference between the points (3,1) and (1,3)

dV=-100(1-3)-100(3-1)=0 V

Answer 2

Question:

The electric field in a region of space is given by  $E = 2x\hat i + 3y\hat j$ (V/m). The magnitude of potential difference between the points (1, 2) & (3,1) is  $= \frac{\alpha}{10}$. Find value of $\alpha$.

Answer:

We find the value of  $\alpha = -35$

Explanation:

Given:   Electric field:  $E=2x\hat i+3y\hat j$

Points-   $(1,2)$ and $(3,1)$

Potential is given by the formula: $dV = - E.dr$

Potential = $dV = - E.dr$

Integrating both sides gives,

$\int dV = -\int_{1}^{3}2xdx -\int_{2}^{1}(3y)dy$

$V=-(3^{2} -1^{2} ) - (\frac{3}{2}( 1^{2}- 2^{2} )$

$V=-(9-1 ) - (\frac{3}{2}(1-4 ))$

$V=-8+\frac{9}{2}$

$V=-\frac{7}{2} = 3.5V$

$V=\frac{\alpha}{10}=-\frac{7 }{2}$

Hence, value of  $\alpha = -35$

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