Two charges of $+ 200 μ C$ and $- 200 μ C$ are placed at the corners B and C of an equilateral triangle ABC of side $0.1 m$ . The force on a charge of $5 μ C$ placed at A is :

1800 N

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

1200 \sqrt{3} N

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

600 \sqrt{3} N

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

900 N

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

Open in App

Solution

The correct option is C $900 N$
$\to$ The vertical components of force cancel out as the magnitude of charges are equal.
Net force experience by A is
$F_{n e t} = \frac{k (200 \times 10^{- 6}) (5 \times 10^{- 6})}{(0 \cdot 1)^{2}} \times C o s (60^{0}) + \frac{k (200 \times 10^{- 6}) (5 \times 10^{- 6})}{(0 \cdot 1)^{2}} (C o s 60)$

$= \frac{(9 \times 10^{9}) (200 \times 10^{- 6}) (5 \times 10^{- 6})}{10^{- 2}}$

$= 9 \times 20 \times 5$
$F_{n e t} = 900 N (^x)$

Suggest Corrections

Similar questions

Q. Charges of

5 μ C

each are placed at the corners of an equilateral triangle of side

10

cm. Then the force on each charge is:-

Q. Point charges of + 3nC and - 3nC are placed at the corners A and of an equilateral triangle ABC each side of which is 0.3 m. What is the force acting on a + 2nc charge placed at C?
(6

\times 10^{- 7}

N parallel to AB)

Q. Point charges +q, +q and -q are placed at the corners A, B and C respectively of an equilateral triangle ABC. The resultant force on '-q' placed at C acts :

Q. 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

Charges of $40 μ C, - 40 μ C$ and $10 μ C$ are placed in air at the corners A, B and C respectively of an equilateral triangle of side $2 c m$ . The resultant force on the charge at C is :

Join BYJU'S Learning Program

Two charges of +200μC and −200μC are placed at the corners B and C of an equilateral triangle ABC of side 0.1m. The force on a charge of 5μC placed at A is :

The correct option is C 900N→ The vertical components of force cancel out as the magnitude of charges are equal.Net force experience by A is Fnet=k(200×10−6)(5×10−6)(0⋅1)2×Cos(600)+k(200×10−6)(5×10−6)(0⋅1)2(Cos60)=(9×109)(200×10−6)(5×10−6)10−2=9×20×5Fnet=900N(^x)

Two charges of $+ 200 μ C$ and $- 200 μ C$ are placed at the corners B and C of an equilateral triangle ABC of side $0.1 m$ . The force on a charge of $5 μ C$ placed at A is :