The point charges 1C, 1C and 1C are placed at the vertices A, B and C of an equilateral triangle of side 1m. Then :
(A) The force acting on the charge at A is $9 \times 10^{9} N$
(B) The electric field strength at A is $9 \times 10^{9} N C^{- 1}$

A is correct but B is wrong

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

B is correct but A is wrong

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

Both A and B are wrong

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

Both A and B are correct

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

Open in App

Solution

The correct option is D Both A and B are wrong
According to question
We have
E in x axis get cancelled
E in y direction = 2 $\times E_{1} c o s θ$ $= 2 \times \frac{k 1}{(1)^{2}} \frac{\sqrt{3}}{2}$
$E = 9 \sqrt{3} \times 10^{9} N$
force = E. q
$= 9 \sqrt{3} \times 10^{9} \times 10$
$= 9 \sqrt{3} \times 10^{9} N C^{- 1}$

Suggest Corrections

Similar questions

Q. ABC is an equilateral triangle of side 1. M. D is the midpoint of BC. Charges of 4C, 4C and 3C are placed at B, C and D respectively. Find the resultant electric intensity at A.

Electric charges of $1 μ C, - 1 μ C$ and $2 μ C$ are placed in air at the corners A, B and C respectively of an equilateral triangle ABC having length of each side 10 cm. The resultant force on the charge at C is $(\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} N - m^{2} / c^{2})$ :

Q. Equal charge

q

are placed at the vertices A and B of an equilateral triangle ABC of side

a

. The magnitude of electric field at point C is

(K = \frac{1}{4 π ϵ_{0}})

Two electric charges of $+ 10^{- 9} C$ and $- 10^{- 9} C$ are placed at the corners A and B of an equilateral triangle ABC side 5cm.The electric intensity at C is

Q. Equal charges,

q

each are placed at the vertices A and B of an equilateral triangle ABC of side

a

. The magnitude of electric intensity at the point C is:

Join BYJU'S Learning Program

The point charges 1C, 1C and 1C are placed at the vertices A, B and C of an equilateral triangle of side 1m. Then :(A) The force acting on the charge at A is 9×109N(B) The electric field strength at A is 9×109NC−1

The correct option is D Both A and B are wrongAccording to questionWe haveE in x axis get cancelledE in y direction = 2 ×E1 cosθ =2×k1(1)2 √32E=9√3×109 Nforce = E. q=9√3×109×10=9√3×109NC−1

The point charges 1C, 1C and 1C are placed at the vertices A, B and C of an equilateral triangle of side 1m. Then :
(A) The force acting on the charge at A is $9 \times 10^{9} N$
(B) The electric field strength at A is $9 \times 10^{9} N C^{- 1}$