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Home » 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 × C is placed at this point, what is the force experienced by the test charge?
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?
CBSE | Class 12 | Electric Charges and Fields | NCERT | Physics
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?

i) Given:

Charges:

qA = 3 µC

qB= –3 µC

Distance between them = 20 cm

Diagrammatic representation of the above question is:

Concept:
 
An electric field is a physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them.
An electric field can be described as a field that surrounds electrically charged particles that attract or repels other charged particles. A system of charged particles is also referred to by the term as a physical field.

Calculation: Evaluating the value of Electric field at point \mathrm{O} caused by +3 \mu C charge

E_{1}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{(O A)^{2}}=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1} \quad along \mathrm{OB}

Where \epsilon_{0}= Permittivity of free space and \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}

Evaluating the value of Electric field at point \mathrm{O} caused by -3 \mu C charge,

E_{2}=\left|\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{-3 \times 10^{-6}}{(O B)^{2}}\right|=\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1} \quad along \mathrm{OB}

\therefore E_{1}+E_{2}=2 \times \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1} \quad along OB

[Since the magnitudes of E_{1} and E_{2} are equal and in the same direction]

    \[\therefore E=2 \times 9 \times 10^{9} \times \frac{3 \times 10^{-6}}{\left(10 \times 10^{-2}\right)^{2}} N C^{-1}=5.4 \times 10^{6} NC ^{-1} \text { along OB }\]

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

ii)

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

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

Let the force experienced by the test charge be \mathrm{F}

Concept: The force acting the charge: F=q E

Calculation: F=1.5\times {{10}^{-9}}\times 5.4\times {{10}^{6}}=8.1 \times 10^{-3} N

Final Answer: The force is directed along line OA because of the attraction between the negative test charge and the charge put at the point and the repelling force between the charge placed at the point. As a result, the force experienced by the test charge is q=8.1 \times 10^{-3} N along \mathrm{OA}.

i

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