Question

Two point charges qa= 5uC and qb= -5°C are located 0.2 m apart in vacuum.
a) What is the electric field at the midpoint 0 on the line AB joining the two charges?
b) If a negative test charge of magnitude 2nC is placed at this point, what is the force experienced by the test
charge? ​

Answer 1

Explanation:

a) 90 *10^5

b)180*10^5

Answer 2

Answer:

a) The electric field at the midpoint O joining the two charges is   $90 \times 10^{5} Vm^{-1}$.

b) The force experienced by a test charge of 2nC is $180\times 10^{-4} N$.

Explanation:

Electric Field:

• It is a physical field that exerts a force on all charged particles in the field. This force could be attractive or repulsive.
• It is a vector quantity denoted by $\vec E$.
• The electrostatic field at a point 'r' due to a charge 'q' at the origin in free space is given by,

                          $\vec E = \frac{q}{4\pi \epsilon_{0} r^{2} } \^r$  where $\epsilon_{0}$ is the permittivity of free space and has the value $8.85 \times 10^{-12} C^{2}N^{-1} m^{-2}$.

• The direction of the electric field at a point is determined by placing a unit positive charge at that point. The direction in which this positive charge is pushed by the field is the direction of an electric field.

Electrostatic Force:

• It is the force experienced by charged particles.
• This force could be attractive or repulsive.
• The force between two stationary charged particles is termed Electrostatic Force. It is also called Coulomb's force.
• This force can be determined by Coulomb's law which states that "The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them".
• Mathematically the force between two charges 'q' and 'Q' separated by a distance 'r' is given by,

                           $F=\frac{qQ}{4\pi \epsilon_{0} r^{2} }$

• The direction of force is in a straight line joining the two charges.

Relation between Electric Field and Electrostatic Force:

                           $\vec F = q\vec E$

Step 1:

Given two point charges $q_{a} = 5 \mu C$ and $q_{b} = -5 \mu C$ separated by a distance of 0.2m in vacuum or free space. First, let's find the electric field at point O (refer to the image).

If a unit positive charge is placed at O, then that charge is repelled by the charge $q_{a}$ and attracted by the charge $q_{b}$. Therefore the electric field due to both the charges on the unit positive charge is in the same direction and hence add up.  

Since the charges $q_{a}$ and $q_{b}$ are of the same magnitude the electric field due to these charges are equal in magnitude too.

Step 2:

Let the magnitude electric field due to one of the charges be E. Then the net electric field at point O will be 2E.

                 $E=\frac{q_{a} }{4\pi \epsilon_{0}r^{2} } \\q_{a} = 5 \mu C , r=0.1m\\E=\frac{5 \times 10^{-6} }{4\pi \epsilon_{0}(0.1^{2})}\\\frac{1}{4\pi \epsilon_{0}} = 9 \times 10^9 \\E = 9 \times 10^9 \times 5 \times 10^-6 \times 10^2 = 45\times 10^5 Vm^-1$

Step 3:

Since the net electric field is 2E. So, we have,

                  $2E=90\times10^5Vm^-1$

Step 4:

To find the force on a test charge of charge 2nC,

                 $F=q(2E)\\F=2\times 10^{-9}\times 90 \times 10^5\\F=180 \times 10^{-4} N$

Therefore,

a) The electric field at the midpoint O joining the two charges is   $90 \times 10^{5} Vm^{-1}$.

b) The force experienced by a test charge of 2nC is $180\times 10^{-4} N$.