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Chapter 13: Problem 79

Two point charges \(+4 q\) and \(+q\) are placed \(30 \mathrm{~cm}\) apart. At what point on the line joining them is the electric field zero? (A) \(15 \mathrm{~cm}\) from charge \(4 q\) (B) \(20 \mathrm{~cm}\) from charge \(4 q\) (C) \(7.5 \mathrm{~cm}\) from charge \(q\) (D) \(5 \mathrm{~cm}\) from charge \(q\)

Short Answer

Expert verified
The electric field is zero at \(x = 20\) cm from the charge \(+4q\). The correct answer is (B) \(20 \mathrm{~cm}\) from charge \(4q\).

Step by step solution

01

Write down the formula for electric field

The electric field created by a point charge is given by the formula: \[E = k\frac{Q}{r^2}\] where E is the electric field, k is the constant \(8.9875 脳 10^9 Nm^2 C^{-2}\), Q is the charge, and r is the distance from the charge.
02

Set up the equation for electric field magnitudes cancellation

At the point where the electric field is zero, the magnitudes of the electric fields created by both the charges are equal, hence: \(E_{4q} = E_q\) Substitute the given values and formula for electric field: \[\frac{k4q}{x^2} = \frac{kq}{(30-x)^2}\]
03

Solve for x

Now we will solve for x, which represents the distance from the 4q charge where the electric field is zero. Solve the equation and cancel out some terms: \[\frac{4}{x^2} = \frac{1}{(30-x)^2}\] Cross-multiply: \[4(30-x)^2 = x^2\] Now, expand and simplify: \[4(900 - 60x + x^2) = x^2\] \[3600 - 240x + 4x^2 = x^2\] \[3x^2 - 240x + 3600 = 0\]
04

Factor and solve the quadratic equation

Factor the quadratic equation: \[3(x^2 - 80x + 1200) = 0\] Dividing the equation by \(3\): \[(x-60)(x-20) = 0\] Solving for x, we have two solutions: \(x = 20\) cm and \(x = 60\) cm. Since the total distance between the charges is 30 cm, we can disregard the 60 cm solution.
05

Write down the final answer

The electric field is zero at \(x = 20\) cm from the charge \(+4q\). So the correct answer is (B) \(20 \mathrm{~cm}\) from charge \(4q\).

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Most popular questions from this chapter

A long string with a charge of \(\lambda\) per unit length passes through an imaginary cube of edge \(a\). The maximum flux of the electric field through the cube will be (A) \(\lambda a / \varepsilon_{0}\) (B) \(\frac{\sqrt{2} \lambda a}{\varepsilon_{0}}\) (C) \(\frac{6 \lambda a^{2}}{\varepsilon_{0}}\) (D) \(\frac{\sqrt{3} \lambda a}{\varepsilon_{0}}\)

The electric \(\vec{E}\) is given by \(\vec{E}=a \hat{i}+b \hat{j}\) (where \(a\) and \(b\) is constant and \(\hat{i}, \hat{j}\) are unit vector along \(x\) and \(y\) axis, respectively), the flux passing through a square area of side \(l\) and parallel to \(y=z\) plane is (A) \(b l^{2}\) (B) \(a l^{2}\) (C) \(\sqrt{\left(a^{2}+b^{2}\right)} l^{2}\) (D) \(\sqrt{\left(a^{2}-b^{2}\right)} l^{2}\)

Two plates are \(2 \mathrm{~cm}\) apart. A potential difference of \(10 \mathrm{~V}\) is applied between them, the electric field between the plates is (A) \(20 \mathrm{~N} / \mathrm{C}\) (B) \(500 \mathrm{~N} / \mathrm{C}\) (C) \(5 \mathrm{~N} / \mathrm{C}\) (D) \(250 \mathrm{~N} / \mathrm{C}\)

An uniform electric field in positive \(x\)-direction exists in a region. Let \(A\) be the origin, \(B\) be the point on the \(x\)-axis at \(x=+1 \mathrm{~cm}\) and \(C\) be the point on the \(y\)-axis at \(y=+1 \mathrm{~cm} .\) The potential at the points \(A, B\), and \(C\) are \(V_{A}, V_{B}\), and \(V_{C}\), respectively, then (A) \(V_{A}V_{B}\) (C) \(V_{A}V_{C}\)

Coulomb's law is applicable to, (A) Point charges (B) Spherical charges (C) Like charges (D) All of these
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