Question:
Two point charges A and B of magnitude \(+8 \times 10^{–6} C\) and \(-8 \times 10^{–6} C\) respectively are placed at a distance d apart. The electric field at the middle point O between the charges is \(6.4 \times 10^{4} NC^{–1}\). The distance ‘d’ between the point charges A and B is:
Two point charges A and B of magnitude \(+8 \times 10^{–6} C\) and \(-8 \times 10^{–6} C\) respectively are placed at a distance d apart. The electric field at the middle point O between the charges is \(6.4 \times 10^{4} NC^{–1}\). The distance ‘d’ between the point charges A and B is:
Updated On: Jul 8, 2024
- 2.0m
- 3.0m
- 1.0m
- 4.0m
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The Correct Option is B
Solution and Explanation
Electric field at P will be E=\(\frac{kq}{(d/2)^2}×2=\frac{8kq}{d^2}\)
So, \(\frac{8×9×10^9×8×10^{−6}}{d^2} =6.4\times10^4\)
So, d = 3 m
\(\therefore ,\) The correct option is (B): 3.0 m
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).