Question

Two point charges qa = +3 µC and qb = -3 µC are located 20 cm apart in vacuum.

Answer 1

Question :-

Two point charges $\sf{q_{a} = +3 \mu\:C \:and }\:\sf{ q_{b} = -3\mu\:C}$ are located 20 cm apart in vacuum .Find force of attraction between them .

Answer :-

$\implies \boxed{ \sf{F = 2.02 \: N}} \:$

To Find :-

→ Force of attraction .

Explanation :-

Given that :

$\red{\boxed{ + }} \: \sf{ q_{a} = +3 \mu\:C} \: , \: q_{b} = - 3 \mu\:C \\ \\ \green{\boxed{ + }} \sf{\: r \: = 20 \: cm \: = 0.2 \: m \: }$

We know that ,

→ Force of attraction between two charges

$\implies \boxed{ \sf{F = \frac{1}{4\pi\epsilon_{o}}\:\frac{q_{a}\: q_{b}}{{r}^{2}}}} \\ \\\because\boxed{ \sf{ \frac{1}{4\pi\epsilon_{o}} =9 \times {10}^{9} } \: , \: 1 \mu = {10}^{ - 6} }\\ \therefore \: \\ \\ \implies \sf{ F = 9 \times {10}^{9} \times \frac{3 \times {10}^{ - 6} \times 3 \times {10}^{ - 6} }{0.2 \times 0.2} } \\ \\ \implies \sf{F = 9 \times {10}^{9} \times \frac{9 \times {10}^{ - 12 + 2} }{4} } \\ \\ \implies \sf{F = \frac{81 \times {10}^{ - 1} }{4} } \\ \\ \implies \sf{ \: F = \frac{81}{40} } \\ \\ \implies \boxed{ \sf{F = 2.02 \: N}}$

Hence ,force of attraction between two given charges is 2.02 Newton .

Answer 2

Correct Question:-

• Two point charges qa = +3 µC and qb = -3 µC are located 20 cm apart in vacuum. What is the electric field at the mid point O of the line AB joining the two charges?

AnswEr:-

• $\bf{5.4\times 10^{6}}$ N/C

Explanation:-

→ Distance between two charges AB = 20 cm

AO= OB= 10 cm

Net electric field at point O= E

Electric field at point O caused by +3μC charge,

E1= 3× 10^-6/4πEo(AO)²

= 3×10^-6/4πE0(10×10^-2)² N/C along OB

→ Magnitude of electric field at point O caused by -3μC charge

E2= |-3×-10^-6/ 4πEo(OB)²|

= 3×10^-6/4πEo(10×10^-2)² N/C along OB

•°• E = E1+ E2

→ 2×[ (9×10^9 × 3×20^-6/(10×10^-2)²]

→ 5 .4 × 10⁶ N/C

$\huge\ddot{\smile}$