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Chapter 11: Problem 1590

Two electric charges \(12 \mu \mathrm{c}\) and \(-6 \mu \mathrm{c}\) are placed \(20 \mathrm{~cm}\) apart in air. There will be a point \(P\) on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of \(P\) from \(-6 \mu c\) charge is ..... (A) \(0.20 \mathrm{~m}\) (B) \(0.10 \mathrm{~m}\) (C) \(0.25 \mathrm{~m}\) (D) \(0.15 \mathrm{~m}\)

Short Answer

Expert verified
The distance of point \(P\) from the \(-6 \mu C\) charge is 0.20 m. The correct answer is (A).

Step by step solution

01

Identify Known and Unknown Variables

The known variables from the problem are the electric charges \(Q_1 = 12 \mu C\) and \(Q_2 = -6 \mu C\) and the distance between them \(d = 20 cm\). The point \(P\) lies outside the region between them, where the electric potential is zero. The unknown variable is the distance \(r\) from charge \(Q_2\) to point \(P\).
02

Write Down the Formula

The formula for the electric potential at a point due to a point charge is \(V = \frac{kQ}{r}\). In this case, we need to consider the electric potentials from both charges and set the total potential to zero. This gives us the equation \(\frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} = 0\).
03

Substitute the Known Variables into the Equation

Substituting the known variables gives \(\frac{kQ_1}{r_1} + \frac{kQ_2}{r_2} = 0\). Substituting \(Q_1 = 12 \mu C\), \(Q_2 = -6 \mu C\), and \(r_2 = r\) (the unknown distance from \(Q_2\) to point \(P\)) into the equation gives: \[ \frac{k \cdot 12}{d + r} - \frac{k \cdot 6}{r} = 0 \] Note that \(d = 20 cm = 0.2 m\) is the distance between the two charges, so \(r_1 = d + r = 0.2m + r\). The \(k\) terms look like they will cancel out in the next step.
04

Simplify the Equation and Solve for the Unknown

Dividing the whole equation by \(k\) and then multiplying by \(r(d + r)\) to eliminate the denominators gives: \[ 12r - 6(d + r) = 0 \] Solving this equation for \(r\) gives: \[ r = \frac{6d}{6} = 0.2m \] Therefore, the distance of point \(P\) from the \(Q_2\) charge is 0.20 m. The correct answer is (A).

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Most popular questions from this chapter

A small conducting sphere of radius \(r\) is lying concentrically inside a bigger hollow conducting sphere of radius \(R\). The bigger and smaller sphere are charged with \(\mathrm{Q}\) and \(\mathrm{q}(\mathrm{Q}>\mathrm{q})\) and are insulated from each other. The potential difference between the spheres will be \(\ldots \ldots\) (A) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(Q / R)]\) (B) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(q / r)-(q / R)]\) (C) \(\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(Q / R)+(q / r)]\) (D) $\left[1 /\left(4 \pi \epsilon_{0}\right)\right][(\mathrm{q} / \mathrm{R})-(\mathrm{Q} / \mathrm{r})]$

The plates of a parallel capacitor are charged up to \(100 \mathrm{~V}\). If $2 \mathrm{~mm}$ thick plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6 \mathrm{~mm}\) the dielectric constant of the plate is (A) 5 (B) 4 (C) \(1.25\) (D) \(2.5\)

A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}\(. The separation between its plates is \)\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectric constant \(\mathrm{K}_{1}=3\) and thickness \(\mathrm{d} / 3\) while the other one has dielectric constant \(\mathrm{K}_{2}=6\) and thickness \(2 \mathrm{~d} / 3\). Capacitance of the capacitor is now (A) \(1.8 \mathrm{pF}\) (B) \(20.25 \mathrm{pF}\) (C) \(40.5 \mathrm{pF}\) (D) \(45 \mathrm{pF}\)

A parallel plate capacitor has the space between its plates filled by two slabs of thickness \((\mathrm{d} / 2)\) each and dielectric constant \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) If \(\mathrm{d}\) is the plate separation of the capacitor, then capacity of the capacitor is .......... (A) $\left[\left(2 \mathrm{~d} \in_{0}\right) / \mathrm{A}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]$ (B) $\left[\left(2 \mathrm{~A} \in_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]$ (C) $\left[\left(2 \mathrm{Ad} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]$ d] \(\left(K_{1}+K_{2}\right)\) (D) \(\left[\left(2 \mathrm{~A} \in_{0}\right) /\right.\)

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.
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