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

AI Mentor

Check Your IQ

Free Expert Demo

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

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 17.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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