Two point charges q1 and q2 of magnitude +10-8 C and -10-8 C respectively are placed 0.1 m apart. Calculate the electric field at point P, 0.1 m away from both the charges.

Question

Two point charges q1 and q2 of magnitude +10-8 C and -10-8 C respectively are placed 0.1 m apart. Calculate the electric field at point P, 0.1 m away from both the charges. (A) 3.6 × 104 N C -1 (B) 4 × 103 N C -1 (C) 6 × 103 N C -1 (D) 9 × 103 N C -1

Tardigrade Admin · Accepted Answer

The point equidistant from both the charges will be on their perpendicular bisector as shown

E_{1 C} = E_{2 C} = \frac{( 9 \times 1 0 ^{9} N m ^{2} C ^{- 2} ) ( 1 0 ^{- 8} C )}{( 0.1 m ) ^{2}}

= 9 \times 1 0^{3} N C^{- 1}

E_{C} = E_{1 C} cos 6 0^{\circ} + E_{2 C} cos 6 0^{\circ}

= 2 (9 \times 1 0^{3} N C^{- 1}) \times \frac{1}{2}

= 9 \times 1 0^{3} N C^{- 1}

Q. Two point charges $q_{1}$ and $q_{2}$ of magnitude $+ 1 0^{- 8} C$ and $- 1 0^{- 8} C$ respectively are placed $0.1 m$ apart. Calculate the electric field at point $P, 0.1 m$ away from both the charges.

$3.6 \times 1 0^{4} N C^{- 1}$

$4 \times 1 0^{3} N C^{- 1}$

$6 \times 1 0^{3} N C^{- 1}$

$9 \times 1 0^{3} N C^{- 1}$

Q. Two point charges q1​ and q2​ of magnitude +10−8C and −10−8C respectively are placed 0.1m apart. Calculate the electric field at point P,0.1m away from both the charges.

3.6×104NC−1

4×103NC−1

6×103NC−1

9×103NC−1

The point equidistant from both the charges will be on their perpendicular bisector as shown E1C​=E2C​=(0.1m)2(9×109Nm2C−2)(10−8C)​ =9×103NC−1 EC​=E1C​cos60∘+E2C​cos60∘ =2(9×103NC−1)×21​ =9×103NC−1

Q. Two point charges $q_{1}$ and $q_{2}$ of magnitude $+ 1 0^{- 8} C$ and $- 1 0^{- 8} C$ respectively are placed $0.1 m$ apart. Calculate the electric field at point $P, 0.1 m$ away from both the charges.

$3.6 \times 1 0^{4} N C^{- 1}$

$4 \times 1 0^{3} N C^{- 1}$

$6 \times 1 0^{3} N C^{- 1}$

$9 \times 1 0^{3} N C^{- 1}$