Two point charges each of 20 microcoulomb are placed 50 cm apart in air

(a) The situation is represented in the given figure. O is the mid-point of line AB.

Distance between the two charges, AB = 20 cm

∴ AO = OB = 10 cm

Net electric field at point O = E

Electric field at point O caused by +3μC charge,

`"E"_1 = (3 xx 10^-6)/(4piin_0("AO")^2) = (3 xx 10^-6)/(4piin_0(10 xx 10^-2)^2) "N"//"C" `along OB

Where,

`in_0` = Permittivity of free space

`1/(4piin_0) = 9 xx 10^9 "Nm"^2"C"^-2`

Magnitude of electric field at point O caused by −3μC charge,

`"E"_1 = (-3 xx 10^-6)/(4piin_0("OB")^2) = (3 xx 10^-6)/(4piin_0(10 xx 10^-2)^2) "N"//"C" `along OB

`therefore "E" = "E"_1 + "E"_2`

= `2xx[(9xx10^9)xx(3xx10^-6)/((10xx10^-2)^2)]` ......[since the value of `"E"_1` and `"E"_2` are same, the value is multiplied with 2]

= 5.4 × 106 N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.

(b) A test charge of amount 1.5 × 10−9 C is placed at mid-point O.

q = 1.5 × 10−9 C

Force experienced by the test charge = F

∴ F = qE

= 1.5 × 10−9 × 5.4 × 106

= 8.1 × 10−3 N

The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A.

Therefore, the force experienced by the test charge is 8.1 × 10−3 N along with OA.

Text Solution

Solution : Here, ` q = +- mu C = +- 20xx10^(-6) C`, <br> `2a = 1 cm = 10^(-2)m, r = 50 cm = (1)/(2) m` <br> As `2 a lt lt r`, therefore, intensity on equatorial line of short dipole is <br> ` E = (1)/(4pi in_(0)) (p)/(r^(3)) = (qxx2a)/(4pi in_(0) r^(3))` <br> `= (9xx10^(9)xx20xx10^(-6)xx10^(-2))/((1//2)^(3))` <br> `E = 1*44xx10^(4) N//C`