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Chapter 11: Problem 1604

Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

Short Answer

Expert verified
The electrical potential will be zero at point \(x=4\). The correct answer is (C).

Step by step solution

01

Write the formula for the potential due to a point charge

The formula to calculate the electrical potential 'V' produced by a single point charge 'Q' at a distance 'r' is: \[V = \frac{kQ}{r}\] Where 'k' is the Coulomb's constant: \(k = 8.9875 × 10^9 Nm^2C^{-2}\). Now, let's find the potential at a point 'x' where the electrical potential will be zero.
02

Define the potential due to two point charges

The net potential at a point x due to the two point charges is the algebraic sum of the potential due to each charge: \(V(x) = V_1(x) + V_2(x)\) The potential due to charge \(q_1 = 2 \mu C\) at point x is: \(V_1(x) = \frac{kq_1}{x}\) and the potential due to charge \(q_2 = -1 \mu C\) at point x is: \(V_2(x) = \frac{kq_2}{(6-x)}\) Since we are looking for the points where the electrical potential is zero, we set the net potential equal to zero: \(V(x) = 0\)
03

Determine the points where the potentials cancel out

Using the formulas from step 2, equate \(V_1(x) + V_2(x) = 0\): \[\frac{kq_1}{x} + \frac{kq_2}{(6-x)} = 0\] Now, substitute the values \(q_1 = 2 \mu C\) and \(q_2 = -1 \mu C\): \[\frac{2k}{x} - \frac{k}{(6-x)} = 0\] Clear the denominators by multiplying through by x(6-x) and simplify: \[2k(6 - x) - kx = 0\] \[k(12 - 3x) = 0\] Since the constant 'k' cannot be zero, the remaining term must be zero: \[12 - 3x = 0\] Solve for x: \[x = 4\] So, due to the two charges, the electrical potential will be zero at point x = 4. This means that the correct answer is (C) x = 4. Note that there may be other points where the electrical potential is zero. However, those points would be hidden in the other choices (A), (B), and (D). Therefore, the correct choice is (C), where x = 4.

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Most popular questions from this chapter

Four equal charges \(\mathrm{Q}\) are placed at the four corners of a square of each side is ' \(\mathrm{a}\) '. Work done in removing a charge \- Q from its centre to infinity is ....... (A) 0 (B) $\left[\left(\sqrt{2} \mathrm{Q}^{2}\right) /\left(\pi \epsilon_{0} \mathrm{a}\right)\right]$ (C) $\left[\left(\sqrt{2} Q^{2}\right) /\left(4 \pi \epsilon_{0} a\right)\right]$ (D) \(\left[\mathrm{Q}^{2} /\left(2 \pi \epsilon_{0} \mathrm{a}\right)\right]\)

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The electric Potential at a point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( is given by \)\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{x} \mathrm{z}^{3}+4\(. The electric field \)\mathrm{E}^{\boldsymbol{T}}$ at that point is \(\ldots \ldots\) (A) $i \wedge\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\mathrm{j} \wedge \mathrm{x}^{2}+\mathrm{k} \wedge 3 \mathrm{xz}^{2}$ (B) $i \wedge 2 \mathrm{xy}+\mathrm{j} \wedge\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\mathrm{k} \wedge\left(3 \mathrm{xy}-\mathrm{y}^{2}\right)$ (C) \(i \wedge z^{3}+j \wedge x y z+k \wedge z^{2}\) (D) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)
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