Three charges are placed as shown in the figure, the resultant force on which of the following charge will be minimum?

F1 Prabhu Ravi 15.04.21 D7

  1. Charge at A
  2. Charge at B
  3. Charge at C
  4. Can't say

Answer (Detailed Solution Below)

Option 2 : Charge at B
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Detailed Solution

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CONCEPT:

The force between multiple charges:

  • Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. 
  • The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.
  • The principle of superposition says that in a system of charges q1, q2, q3,..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges  q2, q3,..., qn. The total force F1 on the charge q1, due to all other charges, is then given by the vector sum of the forces F12, F13, ..., F1n.


\(⇒ \vec{F_1}=\vec{F_{12}}+\vec{F_{13}}+...+\vec{F_{1n}}\)

CALCULATION:

F1 Prabhu Ravi 15.04.21 D7

The force between the charge at A and B is given as,

\(⇒ F_{AB} = k\frac{{{q}\; × \;{q}}}{{{r^2}}}\)

\(⇒ F_{AB} = k\frac{{{q^2}}}{{{r^2}}}=F\)     -----(1)

The force between the charge at B and C is given as,

\(⇒ F_{BC} = k\frac{{{q}\; × \;{q}}}{{{r^2}}}\)

\(⇒ F_{BC} = k\frac{{{q^2}}}{{{r^2}}}=F\)     -----(2)

The force between the charge at A and C is given as,

\(⇒ F_{AC} = k\frac{{{q}\; × \;{q}}}{{{(2r)^2}}}\)

\(⇒ F_{AC} = k\frac{{{q^2}}}{{{4r^2}}}=\frac{F}{4}\)     -----(3)

The charge at B and at C will repel the charge at A, so the resultant force on the charge at A is given as,

F1 Prabhu Ravi 15.04.21 D11

\(⇒ F_A=F_{AC}+F_{AB}\)

\(⇒ F_A=F+\frac{F}{4}\)

\(⇒ F_A=\frac{5F}{4}\)

\(⇒ F_{A} = \frac{{{5kq^2}}}{{{4r^2}}}\)     -----(4)

The charge at A and at B will repel the charge at C, so the resultant force on the charge at C is given as,

F1 Prabhu Ravi 15.04.21 D12

\(⇒ F_C=F_{BC}+F_{AC}\)

\(⇒ F_C=F+\frac{F}{4}\)

\(⇒ F_C=\frac{5F}{4}\)

\(⇒ F_{C} = \frac{{{5kq^2}}}{{{4r^2}}}\)     -----(5)

The charge at A and at C will repel the charge at B, so the resultant force on the charge at B is given as,

F1 Prabhu Ravi 15.04.21 D13

\(⇒ F_B=F_{BC}-F_{AB}\)

⇒ FB = F - F

⇒ FB = 0     -----(6)

  • By equation 4, equation 5, and equation 6 it is clear that the resultant force will be minimum on the charge at B. Hence, option 2 is correct.
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