Chapter 16, Problem 22Q

Given two point charges, Q and 2Q, a distance l apart, is there a point along the straight line that passes through them where E = 0 when their signs are (a) opposite, (b) the same? If yes, state roughly where this point will be.

To determine

Part(a)To Determine:

A point along the straight line that passes through two charges where $E=0$ when the signs of the charges are opposite.

Answer to Problem 22Q

Solution:

The point along the straight line joining the two charges where $E=0$lies outside of the charges and at a distance of $2.41l$ away from the weaker charge and $3.41l$ away from the stronger charge if the charges have opposite signs and placed $l$ distance apart.

Explanation of Solution

Given Info:

The two charges $Q$ and $2Q$ are placed $l$ distance apart.

Formula Used:

The electric field at a point $r$ distance apart from a point charge $q$ is given by,

$E=\frac{kq}{r^2}$

Calculation:

Let $Q$ be negative and $2Q$ be positive. Since the two charges are of opposite signs then the point at which $E=0$ lies outside of the charges.

The electric fields at the point P due to the two charges are in opposite direction.

The point P is at a distance of $x$ from the charge $Q$.

If $E=0$ at P, then $E_Q=E_{2Q}$ at the point P

That is, $\frac{kQ}{x^2}=\frac{k\times 2Q}{{\left(l+x\right)}^2}$

Therefore, $2x^2={\left(l+x\right)}^2$ and $x^2-2lx-l^2=0$

Solving we get, $x=2.41l$

The point P lies at $2.41l$ away from $-Q$ and $3.41l$ away from $2Q$

If two charges having opposite sign are placed at a distance $l$ apart, then the point along the straight line joining the two charges where $E=0$ lies outside of the charges and at a distance of $2.41l$ away from the weaker charge and $3.41l$ away from the stronger charge.

To determine

Part(b)To Determine:

A point along the straight line that passes through two charges where $E=0$ when the signs of the charges are the same.

Answer to Problem 22Q

Solution:

The point along the straight line joining the two charges where $E=0$ lies in between the charges and at a distance of $0.41l$ away from the weaker charge and $0.59l$ away from the stronger charge if the charges have same sign and placed $l$ distance apart.

Explanation of Solution

Given Info:

The two charges $Q$ and $2Q$ are placed $l$ distance apart.

Formula Used:

The electric field at a point $r$ distance apart from a point charge $q$ is given by,

$E=\frac{kq}{r^2}$

Calculation:

Let both $Q$ and $2Q$ are positive. Since the two charges are of same sign then

the point at which $E=0$ lies in between the charges.

The electric fields at the point P due to the two charges are in opposite direction.

The point P is at a distance of $x$ from the charge $Q$.

If $E=0$

at P, then $E_Q=E_{2Q}$ at the point P

That is, $\frac{kQ}{x^2}=\frac{k\times 2Q}{{\left(l-x\right)}^2}$

Therefore, $2x^2={\left(l-x\right)}^2$ and $x^2+2lx-l^2=0$

Solving we get, $x=0.41l$

The point P lies at $0.41l$ away from $Q$ and $0.59l$ away from $2Q$

If two charges having same sign are placed at a distance $l$ apart, then the point along the straight line joining the two charges where $E=0$ lies in between the charges and at a distance of $0.41l$ away from the weaker charge and $0.59l$ away from the stronger charge.