A charged drop is suspended in uniform field of \[3 \times {10^4}\,{\text{V}}{{\text{m}}^{ - 1}}\] so that it neither falls nor rises. The charge on the drop will be:
(Take the mass of the charge \[ = 9.9 \times {10^{ - 15}}\,{\text{kg}}\] and \[g = 10\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\])
A. \[3.3 \times {10^{ - 18}}\,{\text{C}}\]
B. \[3.2 \times {10^{ - 18}}\,{\text{C}}\]
C. \[1.6 \times {10^{ - 18}}\,{\text{C}}\]
D. \[4.8 \times {10^{ - 18}}\,{\text{C}}\]
Answer
642.3k+ views
Hint: The charge neither falls nor rises, for this condition all the forces acting on the drop must be balanced. Check for all the forces acting on the drop and balance them to find the charge on the drop.
Complete Step by step answer: Given,electric field, \[E = 3 \times {10^4}\,{\text{V}}{{\text{m}}^{ - 1}}\]
Mass of the drop, \[M = 9.9 \times {10^{ - 15}}\,{\text{kg}}\]
Acceleration due to gravity, \[g = 10\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\]
Let \[q\] be the charge of the drop.
The gravitational force is acting on the drop in downward direction and since the charged drop is in electric field there will be Coulomb force too, for the charge drop to neither fall nor rise it must be in equilibrium, that is the gravitational force and Coulomb force must be balanced.
The gravitational force on the drop is
\[{F_g} = Mg\]
And Coulomb force is
\[{F_c} = qE\]
For equilibrium, \[{F_c} = {F_g}\]
\[ \Rightarrow qE = Mg\]
Now, putting the values of \[E\], \[M\] and \[g\] from given conditions, we get
\[\Rightarrow q \times \left( {3 \times {{10}^4}} \right) = \left( {9.9 \times {{10}^{ - 15}}} \right) \times 10 \\
\Rightarrow q = \dfrac{{\left( {9.9 \times {{10}^{ - 15}}} \right) \times 10}}{{\left( {3 \times {{10}^4}} \right)}} \\
\Rightarrow q = 3.3 \times {10^{ - 18}}\,{\text{C}} \\ \]
Therefore, the charge on the drop is \[3.3 \times {10^{ - 18}}\,{\text{C}}\]
Hence, the correct answer is option (A) \[3.3 \times {10^{ - 18}}\,{\text{C}}\]
Note: If there more than two forces, then find the net force acting on the body. For a body to be in equilibrium the net force on the body must be zero. Use this condition for solving problems where there are more than two forces acting on the system.
Complete Step by step answer: Given,electric field, \[E = 3 \times {10^4}\,{\text{V}}{{\text{m}}^{ - 1}}\]
Mass of the drop, \[M = 9.9 \times {10^{ - 15}}\,{\text{kg}}\]
Acceleration due to gravity, \[g = 10\,{\text{m}}{{\text{s}}^{{\text{ - 2}}}}\]
Let \[q\] be the charge of the drop.
The gravitational force is acting on the drop in downward direction and since the charged drop is in electric field there will be Coulomb force too, for the charge drop to neither fall nor rise it must be in equilibrium, that is the gravitational force and Coulomb force must be balanced.
The gravitational force on the drop is
\[{F_g} = Mg\]
And Coulomb force is
\[{F_c} = qE\]
For equilibrium, \[{F_c} = {F_g}\]
\[ \Rightarrow qE = Mg\]
Now, putting the values of \[E\], \[M\] and \[g\] from given conditions, we get
\[\Rightarrow q \times \left( {3 \times {{10}^4}} \right) = \left( {9.9 \times {{10}^{ - 15}}} \right) \times 10 \\
\Rightarrow q = \dfrac{{\left( {9.9 \times {{10}^{ - 15}}} \right) \times 10}}{{\left( {3 \times {{10}^4}} \right)}} \\
\Rightarrow q = 3.3 \times {10^{ - 18}}\,{\text{C}} \\ \]
Therefore, the charge on the drop is \[3.3 \times {10^{ - 18}}\,{\text{C}}\]
Hence, the correct answer is option (A) \[3.3 \times {10^{ - 18}}\,{\text{C}}\]
Note: If there more than two forces, then find the net force acting on the body. For a body to be in equilibrium the net force on the body must be zero. Use this condition for solving problems where there are more than two forces acting on the system.
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