Question

A right isosceles triangle of side a has charges q, +2q and −q arranged on
its vertices, as shown in figure below:
What is the electric field at point P, midway between the line connecting
the +q and −q charges? Give the magnitude and direction of the electric
field.

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given info : A right isosceles triangle of side a has charges q, +2q and −q arranged on

its vertices.

To find : the electric field at point, P midway between the line connecting the +q and -q charges, is...

solution : see diagram,

electric field due to charge placed at A, E = kq/(a/√2)² = 2kq/a² (along AC)

in vector form,

E = 2kq/a² [cos45° i - sin45° j ]

= √2kq/a² i - √2kq/a² j

electric field due to charge placed at C, E = 2kq/a² (along AC)

in vector form, E = √2kq/a² i - √2kq/a² j

electric field due to charge placed at B, E' = 4kqa²

in vector form, E' = 4kq/a² [cos45° i + sin45° j]

= 2√2kq/a² i + 2√2kq/a² j

now resultant electric field, Enet = E + E + E'

= √2kq/a² i - √2kq/a² j + √2kq/a² i - √2kq/a² j + 2√2kq/a² i + 2√2kq/a² j

= 4√2kq/a² i

Therefore the magnitude of electric field is 4√2kq/a² and direction of electric field is along positive x - axis.