1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Four positive charges (2√2−1)Q are arranged at the four corners of a square. Another charge q is placed at the centre of the square. Resulting field acting on each corner will be zero if q is

A
7Q4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
4Q7
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Q
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(2+1)Q
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A −7Q4 Given: Q1=(2√2−1)Q So, E1=kQ1a2 E2=kQ1(√2a)2 E3=kq(a√2)2 For net electric field to be zero, →E3+→E1+→E1+→E2=0 ⇒→E3 should balance electric field due to other charges, hence q should be negative in nature. Thus →E3 acts towards charge q ⇒|→E3|=√E21+E21+|→E2| [From figure] ⇒2kqq2=√(kQ1a2)2+(kQ1a2)2+kQ12a2 ⇒2kqa2=√2×kQ1a2+kQ12a2 ⇒2q=Q1[2√2+12] ⇒q=Q1[2√2+14] ⇒q=2√2+14×(2√2−1)Q As q is negative in nature, ⇒q=−7Q4 Hence, option (a) is correct.

Suggest Corrections
35
Join BYJU'S Learning Program
Join BYJU'S Learning Program