An electric dipole shown in the figure. Work done to move a charge particle of \(1\mu C\) from point Q to P is \(x \times 10^{-7} J\), then the value of \(x\) is:
Show Hint
In dipole problems, same \(r\) $\Rightarrow$ focus on \(\cos\theta\) difference.
Step 1: Understanding the Concept:
Work done in moving a charge \(q\) is \(W = q(V_{\text{final}} - V_{\text{initial}})\). Here we need the potential of a dipole at points P and Q. Step 2: Key Formula or Approach:
Potential due to dipole \(V = \frac{kp \cos\theta}{r^{2}}\).
Dipole moment \(p = q \times 2l\). : Detailed Explanation:
Dipole moment \(p = (2 \times 10^{-6}\text{ C}) \times (1 \times 10^{-9}\text{ m}) = 2 \times 10^{-15}\text{ Cm}\).
Distance \(r = 1\text{ cm} = 0.01\text{ m}\) for both points.
For P: \(\theta = 60^{\circ}\).
\[ V_{P} = \frac{(9 \times 10^{9})(2 \times 10^{-15}) \cos 60^{\circ}}{(0.01)^{2}} = 0.09\text{ V} \]
For Q: \(\theta = 120^{\circ}\) (relative to the positive direction of the dipole).
\[ V_{Q} = \frac{(9 \times 10^{9})(2 \times 10^{-15}) \cos 120^{\circ}}{(0.01)^{2}} = -0.09\text{ V} \]
Potential difference \(\Delta V = V_{P} - V_{Q} = 0.09 - (-0.09) = 0.18\text{ V}\).
Work done \(W = (1 \times 10^{-6}) \times 0.18 = 1.8 \times 10^{-7}\text{ J}\).
Comparing with \(x \times 10^{-7}\), we get \(x = 1.8\). Step 3: Final Answer:
The value of \(x\) is \(1.8\).
