Question

Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apartin vacuum. (a) What is the electric field at the midpoint O of the line AB joiningthe two charges? (b) If a negative test charge of magnitude 1.5 × 10–9 C is placed atthis point, what is the force experienced by the test charge?

Answer 1

a)

Given: The electric charges q A =3 μCand q B =−3 μC are located 20 cm apart.

Consider the figure below,

The electric field produced by a charge q at a distance ris given as,

E( r )= 1 4π ε 0 q r 2   r ^ .(1)

Since O is the midpoint of line AB ,

OB=OA=0.1 m.

By substituting the values for charge q A in equation (1), we get

E 1 =9× 10 9 × 3× 10 −6 ( 0.1 ) 2 =2.7× 10 6  N/C

By substituting the values for charge q B in equation (1), we get

E 2 =9× 10 9 × 3× 10 −6 ( 0.1 ) 2 =2.7× 10 6  N/C

Total electric field due to both charges is given as,

E Total = E 1 + E 2

By substituting the values in the above equation, we get

E Total =( 2.7+2.7 )× 10 6 =5.4× 10 6  N/C

Thus, the total electric field at the midpoint O is 5.4× 10 6  N/C.

b)

Given: A negative charge of magnitude 1.5× 10 −9  C is placed at O.

The electrostatic force experienced by the test charge is given as,

F( r )=q×E( r ).

By substituting the given values in the above equation, we get

F=1.5× 10 −9 ×5.4× 10 6 =8.1× 10 −3  N

Test charge is negative, so negative charge at point B will repel and positive charge at point A will attract.

Thus, the magnitude of force will be 8.1× 10 −3  N along the direction OA.