AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Write the expression of electric field at A, in the vector form.
EA={

kp

(12+22+32)3∕2

} (

^

i

+2

^

j

+3

^

k

)
=

kp

(14)3∕2

(

^

i

+2

^

j

+3

^

k

)
Write the expression of electric field at B, in the vector form.
EB={

kp

(12+12+(−1)2)3∕2

} (

^

i

+

^

j

−

^

k

)
=

kp

(3)3∕2

(

^

i

+

^

j

−

^

k

)
Write the expression of electric field at C, in the vector form.
EC={

kp

(22+22+22)3∕2

} (2

^

i

+2

^

j

+2

^

k

)
=

kp

(12)3∕2

(2

^

i

+2

^

j

+2

^

k

)
So,

→

E

A⊥

→

E

B=0
And |

→

E

B|=4|

→

E

C|