Three point charges of magnitude 5μC, 0.16μC and 0.3μC are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3cm, BC=32 cm and CA = 3 cm and point A is the right angle corner. Charge at point A experiences _______ N of electrostatic force due to the other two charges.

F1=k×5×0.3×10−129×10−4=9×109×5×0.3×10−129×10−4=1.5×10=15NF2=9×109×5×0.16×10−129×10−4=8N force experienced by charge at A=F12+F22=152+82=289=17N

Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Three point charges of magnitude $5 μC, 0 .16 μC and 0 .3 μC$ are located at the vertices A, B, C of a right angled triangle whose sides are AB = 3cm, $BC = 3 \sqrt{2} cm$ and CA = 3 cm and point A is the right angle corner. Charge at point A experiences _______ N of electrostatic force due to the other two charges.

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 17.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} F_{1} = \frac{k \times 5 \times 0 .3 \times 10^{- 12}}{9 \times 10^{- 4}} \\ = \frac{9 \times 10^{9} \times 5 \times 0 .3 \times 10^{- 12}}{9 \times 10^{- 4}} \\ = 1 .5 \times 10 = 15 N \\ F_{2} = \frac{9 \times 10^{9} \times 5 \times 0 .16 \times 10^{- 12}}{9 \times 10^{- 4}} = 8 N \\ force experienced by charge at A = \sqrt{F_{1}^{2} + F_{2}^{2}} \\ = \sqrt{15^{2} + 8^{2}} \\ = \sqrt{289} = 17 N \end{array}$