Question:

Two point charges +10−7C + 10^{-7} C and −10−7C - 10^{-7} C are placed at A A and B,20 cm B, 20 \,cm apart as shown in the figure. Calculate the electric field at C, 20 cm C, \,20\, cm apart form both A A and B B .

Updated On: Jun 14, 2022
  • 1.5×10−5N/C 1.5 \times 10^{-5} N/C
  • 2.2×104N/C 2.2 \times 10^{4} N/C
  • 3.5×106N/C 3.5 \times 10^{6} N/C
  • 3.0×105N/C 3.0 \times 10^{5} N/C
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

The magnitude of each electric field vector at point CC due to charge q1q_1 and q2q_2
E=14πε0⋅qr2E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}
E=9×109×10−7(20×10−2)2E = \frac{9 \times 10^9 \times 10^{-7}}{(20 \times 10^{-2})^2}
=2.2×104 N/C = 2.2 \times 10^4 \,N/C
Was this answer helpful?
0
0

Top Questions on Electric Field

View More Questions

Questions Asked in AMUEEE exam

View More Questions

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∣q1q2∣r2F=k\dfrac{|q_{1}q_{2}|}{r^{2}}

While substituting q2 as 1, electric field becomes:

 E=k∣q1∣r2E=k\dfrac{|q_{1}|}{r^{2}}

SI unit of Electric Field is V/m (Volt per meter).