Four point charges $q_{A} = 2 μ C, q_{B} = - 5 μ C, q_{C} = 2 μ C$ and $q_{D} = - 5 μ C$ are located at the corners.
A, B, C, D of a square ABCD of side 10 cm. What is the force on a charge of $1 μ C$ placed at the centre of the square?

2 \times 10^{3} N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \times 10^{3} N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8 \times 10^{3} N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Zero

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C Zero
From principle of super position.
$F_{N e t} = F_{A} + F_{B} + F_{C} + F_{D}$ .
Notice that $F_{A} = - F_{C}$ and $F_{D} = - F_{B}$ because of symmetry.
Thus, $F_{N e t} = 0$

Suggest Corrections

Similar questions

q_{A} = 2 μ C

q_{B} = 5 μ C

q_{C} = 2 μ C

q_{D} = 5 μ C

1 μ C

q_{A} = 2 μ C, q_{B} = 5 μ C, q_{C} = 2 μ C, a n d q_{D} = 5 μ C

10 c m

q_{A} = μ c, q_{B} = - 5 μ c q_{c} = 2 μ c a n d q_{D} = - 5 μ c

5 μ C, - 2 μ C, 3 μ C

- 9 μ C

1

0.9 m

- 2 μ C, + 3 μ C, - 4 μ C

+ 5 μ C

Join BYJU'S Learning Program