NEET · Physics · STD 12 - 2. Electric potential and capacitance
An electric dipole is placed as shown in the figure.The electric potential (in \(10^2\,V\) ) at point \(P\) due to the dipole is \(\left(\epsilon_0=\right.\) permittivity of free space and \(\left.\frac{1}{4 \pi \epsilon_0}=K\right)\) :
- A \(\left(\frac{8}{3}\right) qK\)
- B \(\left(\frac{3}{8}\right) qK\)
- C \(\left(\frac{5}{8}\right) qK\)
- D \(\left(\frac{8}{5}\right) qK\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{3}{8}\right) qK\)
Step-by-step Solution
Detailed explanation
\(v =\frac{ Kq }{2 \times 10^{-2}}-\frac{ Kq }{8 \times 10^{-2}}\) \(= Kq \left[\frac{3}{8}\right] \times 10^2\)
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