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

Question

Two point charges + 10-7 C and - 10-7 C are placed at A and B, 20 cm apart as shown in the figure. Calculate the electric field at C, 20 cm apart form both A and B . <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/633741d3c77203daea6f327d1f71bf99-.png" /> (A)  1.5 × 10-5 N/C  (B)  2.2 × 104 N/C  (C)  3.5 × 106 N/C  (D)  3.0 × 105 N/C

Tardigrade Admin · Accepted Answer

The magnitude of each electric field vector at point C due to charge q1 and q2 E = (1/4π ε0) ⋅ (q/r2) E = (9 × 109 × 10-7/(20 × 10-2)2) = 2.2 × 104 N/C

Q. Two point charges $+ 1 0^{- 7} C$ and $- 1 0^{- 7} C$ are placed at $A$ and $B, 20 c m$ apart as shown in the figure. Calculate the electric field at $C, 20 c m$ apart form both $A$ and $B$ .

$1.5 \times 1 0^{- 5} N / C$

$2.2 \times 1 0^{4} N / C$

$3.5 \times 1 0^{6} N / C$

$3.0 \times 1 0^{5} N / C$

The magnitude of each electric field vector at point $C$ due to charge $q_{1}$ and $q_{2}$
$E = \frac{1}{4 π ε _{0}} \cdot \frac{q}{r ^{2}}$
$E = \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 7}}{( 20 \times 1 0 ^{- 2} ) ^{2}}$
$= 2.2 \times 1 0^{4} N / C$

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

1.5×10−5N/C

2.2×104N/C

3.5×106N/C

3.0×105N/C