Four charges as shown in the figure are placed at the corners of a square ABCD. The electric intensity on a unit charge kept at the centre O of the square is
Let the small unit charge +Q be kept at centre. The field EA due to charge +q at A on unit charge Q at centre is
|EA|=|‌
1
4πε0
‌
qQ
OA2
|,‌ along ‌OC Similarly, |EB|=|‌
1
4πε0
‌
qQ
OC2
|, along OA Since, OA=OC, therefore |EA|=|EB| Therefore, resultant field due to EA and EB is E1=‌‌√|EA|2+|EB|2+2|EA||EB|cos‌180∘ =‌‌√2|EA|2+2|EA|2(−1)=0 [∵cos‌180∘=−1] Now, field EB due to charge +2q at B on charge +Q at centre O is |EB|=|‌
1
4πε0
‌
2qQ
OB2
|,‌ along ‌OD and field ED is given by |ED|=|‌
1
4πε0
‌
−2q×Q
OD2
|‌, along ‌OD Therefore, resultant field due to EB and ED is E2‌‌=√|EB|2+|ED|2+2|EB||ED|cos‌0∘ ‌‌=√EB2+EB2+2EB2(1) [∵OB=OD⇒EB=ED] ‌‌=√2EB2+2EB2‌‌[∵cos‌0∘=1] ⇒‌‌E2=2EB ∴ The resultant electric field at point O is along BD.