# Question: The electric field due to this combination of charges can be zero

The electric field due to this combination of charges can be zero

A) Only in region 1

B) Only in region 2

C) Only in region 3

D) in both regions 1 and 3

Transcribed text From Image:
Contents

Answer:

## General guidance

Concepts and reason

The concept used to solve this problem is electric field.

Find the electric field in the three regions. Fundamentals

The electric field is a vector field associated with every point around a point charge. It is the electric force per unit charge.

Electric field at a point is the electric force experienced by a unit charge when placed at that point.

Expression for the electric field is,

E=kqr2E = k\frac{q}{{{r^2}}}E=kr2q​

Here EEE is the electric field, kkk is a constant, qqq is the source charge that produces the electric field, and rrr is the distance between the source charge and the field point.

The magnitude of the charge q1{q_1}q1​ is 4×10−6C4 \times {10^{ – 6}}\,{\rm{C}}4×10−6C .

The magnitude of the charge q2{q_2}q2​ is −2×10−6C- 2 \times {10^{ – 6}}\,{\rm{C}}−2×10−6C . Show more First Step | All Steps | Answer Only

## Step-by-step

### Step 1 of 2

REGION 1:

This region is closer to the charge q1=4×10−6C{q_1} = 4 \times {10^{ – 6}}\,{\rm{C}}q1​=4×10−6C than q2=−2×10−6C{q_2} = – 2 \times {10^{ – 6}}\,{\rm{C}}q2​=−2×10−6C . The electric field is expressed as E=k(q/r2)E = k\left( {q/{r^2}} \right)E=k(q/r2) . EEE is directly proportional to the magnitude of the source charge. The magnitude of the field produced by the charge q1{q_1}q1​ is more in this region than the opposing field produced by the charge q2{q_2}q2​ . Hence there will be a net nonzero electric field in this region.

REGION 2:

This region is located between the two charges q1{q_1}q1​ and q2{q_2}q2​ . Since the charge q1{q_1}q1​ is positive and charge q2{q_2}q2​ is negative, the electric field produced by these charges gets added up. Therefore, the field cannot become zero in this region.

REGION 3:

This region is located near the charge q2{q_2}q2​ and it is located away from the charge q1{q_1}q1​ . The electric field produced by the charge q1{q_1}q1​ acts towards the right and the field produced by the charge q2{q_2}q2​ acts towards left.

Since the field is indirectly proportional to the square of the distance and directly proportional to the magnitude of charge, both fields cancel each other in this region. The electric field produced by the charge q2{q_2}q2​ is balanced by the field due to the charge q1{q_1}q1​ .

Therefore, net field is zero.Explanation | Common mistakes | Hint for next step

The electric field produced by a positive charge will be directed away from the charge and the field produced by a negative charge will be pointed towards the charge.

The incorrect expression for the electric field is E=k(q/r)E = k\left( {q/r} \right)E=k(q/r) . Hence the correct expression is E=k(q/r2)E = k\left( {q/{r^2}} \right)E=k(q/r2) .

Use the assumptions made in the previous step to find the correct option.

### Step 2 of 2

The incorrect options are,

A)Only in region 1

B)Only in region 2

D)In both regions 1 and 3

Therefore, the correct option is,

C)Only in region 3.

The correct option is C) only in region 3.
Explanation | Common mistakes

The electric field cannot be zero in both regions 1 and 3 despite they are located at equal distance from the centre respectively. This is because the magnitude of the one charge is greater than the magnitude of the other.

The incorrect option is A) only in region 1. Hence the correct option is C) only in region 3.