Two charges `10 muC and - 10 muC` are placed at points `A. and B` separated by a distance of 10 cm. Find the electric. field at a point P on the perpendicular bisector of `AB` at. a distance of 12 cm from its middle point.

Explanation

The situatin is shown in figure The distance
.`AP =BP= sqrt((5cm)^2 + (12cm)^2 ) = 13cm`
. The field at the point P due to the charge `10 muC` is
`E_A =(10muC/4pi epsilon (13cm)^2 = ((10 xx 10(-6) xx (9xx10^9 Nm^2 C(-2)) (169 xx 10(-4) m^2`
`= 5.3 xx 10 ^6 NC(-1).`
This field is along AP. The field due to `-10 muC at P` is
. `E_B = 5.3 x 10 ^6 NC(-1)` along PB. as `E_A and E_B` are equal
. in magnitude, the resultant will bisect the angle between
. the two. The geometry of the figure shows that this
. resultant is parallel to the base AB. The magnitude of
. the resultant field is
`E =E_A cos theta + E_B cos theta`
`= 2 xx (5.3 xx 10^6 NC(-1) xx 5/13`
`= 4.1 x 10^6 NC(-1)`
. If a given charge distribution is continuous, we can
. use the technique of intergration to find the resultant.
electric field at a point. A small element dQ is chosen.
in the distributin and the field `d vec E` due to dQ is
. calculated. The resultant field is then calculated by
. intergratin the components of `d vec E` under proper limits
.