Test Index

Electrostatics

Section: Physics

Question : 32 of 32

Marks: +1, -0

(30 \overline{i}+40 j) \mathrm{NC}^{-1}

(1 \mathrm{m}, 2 \mathrm{m})

Solution:

To find the electric potential at the point

(1 \mathrm{m}, 2 \mathrm{m})

, we begin with the given electric field and displacement vector:

\;E = (30 \hat{i} + 40 \hat{j}) \mathrm{N/C}

\;d = (1 \hat{i} + 2 \hat{j}) \mathrm{m}

The formula for electric potential

V

with respect to a chosen reference point is given by:

V = -E \cdot r

Applying the formula, we compute the dot product:

V = -(30 \hat{i} + 40 \hat{j}) \cdot (1 \hat{i} + 2 \hat{j})

Break it down as follows:

V \;= -(30 \cdot 1 + 40 \cdot 2)

V \;= -(30+80)

V \;= -110 \mathrm{V}

Thus, the electric potential at the point

(1 \mathrm{m}, 2 \mathrm{m})

is -110 V .

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