Determine the electric potential at the mid point of line two charges 2×10-⁶c & -1×10-⁶c placed in a vaccum 10cm apart
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Explanation:
Two point charges qA = 3mu C and qB = - 3 mu C 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^-9C is placed at this point, what is the force experienced by the test charge?
Answer · 118 votes
(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 + 3 C charge, E1 = 3 × 10^-6/4pi∈0(AO)^2 = 3 × 10^-6/4pi∈0(10 × 10^-2)^2N/C along OBWhere, ∈0 = Permittivity of free space 1/4pi∈0 = 9 × 10^9Nm^2C^-2 Magnitude of electric field at point O caused by - 3 C charge, E2 = |-3 × 10^-6/4pi∈0(OB)^2 | = 3 × 10^-6/4pi∈0(10 × 10^-2)^2N/C along OB E = E1 + E2 = 2 × [(9 × 10^9) × 3 × 10^-6/(10 × 10^-2)^2 ] [As E1 = E2 , the value is multiplied with 2] = 5.4 × 10^6 N/ C along OBTherefore, the electric field at mid - point O is 5.4 × 10^6 N/ C along OB.(b) A test charge of amount q = 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 × 10^6 = 8.1 × 10^-3 N The force is along line OA. This is because the negative test charge is repelled by the charge placed at p…