Question

Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum. (i) What is the electric field at the midpoint O of the line AB joining the two charges? (ii) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?

Answer 1

$\bigstar$ Question:

Two point charges $\sf qA=3\: \mu C$ and $\sf qB=-3 \: \mu C$ are located 20 cm apart in a vacuum.

(i) What is the electric field at the midpoint O of the line AB joining the two charges?

(ii) If a negative test charge of magnitude $\sf 1.5 \times 10^{-9} \:C$ is placed at this point, what is the force experienced by the test charge?

$\bigstar$ Given:

• Distance between two charges, AB = 20 cm
• Total electric field at the centre is (Point O) = E
• Electric field at point O caused by = $\sf +3 \: \mu C$

$\bigstar$ To Find:

• The electric field at the midpoint O of the line AB joining the two charges.
• Force experienced by the test charge if a negative test charge of magnitude $\sf 1.5 \times 10^{-9}C$

$\bigstar$ Solution:

⚫ CASE I:

According to the given information:

Distance between two charges, AB = 20 cm

Therefore, AO = OB = 10 cm

Total electric field at the centre is (Point O) = E

Electric field at point O caused by $\sf +3 \: \mu C$ charge,

$\sf E_1=\dfrac{1}{4 \pi \epsilon_0} .\dfrac{3 \times 10^{-6}}{(AO)^{2}} =\dfrac{1}{4 \pi \epsilon_0} .\dfrac{3 \times 10^{-6}}{(10 \times 10^{-2})^{2}}\: NC^{-1}$  along OB

Where, $\epsilon_0=$ Permittivity of free space and $\sf \dfrac{1}{4 \pi \epsilon_0} =9 \times 10^{9}\: Nm^{2}\: C^{-2}$

Therefore,

Electric field at point O caused by $\sf -3 \: \mu C$ charge,

$\sf E_2 = \bigg[\dfrac{1}{4 \pi \epsilon_0} .\dfrac{-3 \times 10^{-6}}{(OB)^{2}} \bigg]=\dfrac{1}{4 \pi \epsilon_0} .\dfrac{3 \times 10^{-6}}{(10 \times 10^{-2})^{2}} \: NC^{-1}$ along OB

∴ $\sf E_1+ E_2= 2 \times \dfrac{1}{4 \pi \epsilon_0} .\dfrac{3 \times 10^{-6}}{(10 \times 10^{-2})^{2}} \: NC^{-1}$ along OB

Since the magnitudes of $E_1$ and $E_2$ re equal and in the same direction,

∴ $\sf E=2 \times 9 \times 10^{9} \times \dfrac{3 \times 10^{-6}}{(10 \times 10^{-2})^{2}} \: NC^{-1}$

$\implies \sf 5.4 \times 10^{6} \: NC^{-1}$ along OB

Therefore, the electric field at mid – point O is $\sf 5.4 \times 10^{6} \: NC^{-1}$ along OB

⚫ CASE II:

A test charge with a charge potential of $\sf 1.5 \times 10^{-9}\: C$ is placed at mid – point O.

$\sf q=1.5 \times 10^{-9} \: C$

Let the force experienced by the test charge be  F

Therefore, F = qE

$\implies \sf 1.5 \times 10^{-9} \times 5.4 \times 10^{6}$

$\implies \sf 8.1 \times 10^{-3}\: N$

The force is directed along line OA because the negative test charge is attracted towards point A and is repelled by the charge placed at point B. As a result, the force experienced by the test charge is $\sf q=8.1 \times 10^{-3} \: N$ along OA.

$\bigstar$ Points to Note:

➤ Electrostatics deals with the study of forces, fields and potentials arising from static charges.

➤ The three fundamental properties of the electric charge are Quantization, Additive and Conservation.

➤ Coulomb’s Law consists of constant terms which are also called a constant of proportionality and is represented by ‘k’

➤ There are two kinds of electrification wherein like charges repel and unlike charges attract each other.  The property that differentiates these 2 kinds of charges is called the polarity of charge.

Answer 2

$\huge{\colorbox{pink}{ Answer }}$

(a) The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm

∴ AO = OB = 10 cm

Net electric field at point O = E

Electric field at point O caused by +3μC charge,

Where, 0 = Permittivity of free space and 1/4πεo=9×109 Nm2C−2

Therefore,

Magnitude of electric field at point O caused by −3μC charge,

[ Since the magnitudes of E1 and E2 are equal and in the same direction ]

Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.

(b) A test charge of amount 1.5 × 10−9 C is placed at mid – point O.

q = 1.5 × 10−9 C

Force experienced by the test charge = F

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A. Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.