Question

ABCD is a square of side 0.2 m. Charges 2 nC, 4 nC & 6 nC are placed at the corners A, B & C respectively. Calculate magnitude and direction of the resultant electric field at D.​

Answer 1

Answer:

so nhi aati kya abcd

Explanation:

aese hoti hai dekho abcdefghijklmnopqrstuvwxyz

Answer 2

Answer:

AC=BD=

(0.2)

2

+(0.2)

2

=

2

×0.2=0.28m

∴AO=BO=CO=

2

0.28m

=0.14m

V i.e, Potential at O;V

o

=

4πε

0

1

[

AO

q

1

+

BO

q

2

+

CO

q

3

]

V i.e, Potential at D;V

D

=

4πε

0

1

[

AD

q

1

+

BD

q

2

+

CD

q

3

]

Work done =q[V

o

−V

D

]

=

4πε

0

2×10

−9

[

0.14

2×10

−9

+

0.14

4×10

−9

+

0.14

8×10

−9

−

0.2

2×10

−9

−

0.28

4×10

−9

−

0.2

8×10

−9

]

On solving, we get

W=2×10

−9

×2×10

9

[

0.14

14×10

−9

−

0.2

10×10

−9

−

0.28

4×10

−9

]

=4×[10

−7

−0.5×10

−7

−0.14×10

−7

]

=5.40×10

−7

Joules.

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