An electric dipole is placed as shown in the figure.

The electric potential (in 102 V ) at point P due to the dipole is $(?_{0} =$ permittivity of free space and $\frac{1}{4 π ?_{0}} = K)$ :

Option 1 -

$(\frac{8}{3}) q K$

Option 2 -

$(\frac{3}{8}) q K$

Option 3 -

$(\frac{5}{8}) q K$

Option 4 -

$(\frac{8}{5}) q K$

2 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Raj Pandey | Contributor-Level 9

3 weeks ago

Correct Option - 2

Detailed Solution:

$V_{P} = V_{q} + V_{- q}$

$\begin{array}{r} = (\frac{K q}{5 - 3} + \frac{K (- q)}{5 + 3}) \times 10^{2} \\ = (\frac{K q}{2} - \frac{K q}{8}) \times 10^{2} \\ = (\frac{3 K q}{8}) \times 10^{2} \end{array}$

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An electric dipole is placed as shown in the figure.

The electric potential (in 102 V ) at point P due to the dipole is $(?_{0} =$ permittivity of free space and $\frac{1}{4 π ?_{0}} = K)$ :

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An electric dipole is placed as shown in the figure. The electric potential (in 102 V ) at point P due to the dipole is ? 0 = permittivity of free space and 1 4 π ? 0 = K :

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An electric dipole is placed as shown in the figure.

The electric potential (in 102 V ) at point P due to the dipole is $(?_{0} =$ permittivity of free space and $\frac{1}{4 π ?_{0}} = K)$ :