# Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? - Physics

Numerical

Two point charges qA = 3 μC and qB = −3 μC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?

#### Solution

(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.

Concept: Electric Field - Physical Significance of Electric Field
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Chapter 1: Electric Charges and Fields - Exercise [Page 46]

#### APPEARS IN

NCERT Physics Class 12
Chapter 1 Electric Charges and Fields
Exercise | Q 1.8 | Page 46
NCERT Physics Class 12
Chapter 1 Electric Charge and Fields
Exercise | Q 8 | Page 46
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