Four equal point charges each of \[16\,\mu C \] are placed on the four corners of a square of side \[{\text{0}}{\text{.2}}\,{\text{m}}\]. Calculate the force on any of the charges.
Answer
584.4k+ views
Hint: We are asked here to find the force on any of the charges on the corners of the square. Here, you will need to use coulomb’s law to find the force between two charges. Use this law to find the force exerted on any one point charge due to the other three charges. Then find the resultant force exerted on a point charge.
Complete step by step answer:
Given, charge on four corners of a square is, \[Q = 16\,\mu C = 16 \times {10^{ - 6}}\,{\text{C}}\]
Side of the square, \[d = {\text{0}}{\text{.2}}\,{\text{m}}\]
Here, we will use coulomb’s law, which states force between two charges is given as,
\[F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}\] ……………(i)
where \[{q_1}\] and \[{q_2}\] are the two charges, \[r\] is the distance between the centre of two charges and \[k\] is the proportionality constant.
Let ABCD be the square with side \[d = {\text{0}}{\text{.2}}\,{\text{m}}\] and point charge \[Q = 16\,\mu C\] on each corner of the square.
We draw the diagram with the given information.
We are asked to calculate the force on any of the charges. The force on any of the charges will be the same.
Let us calculate the force on point A. We will now see the force exerted by the other three points on point A.Force exerted by point B on point A, using coulomb’s law from equation (i) we get,
\[{F_{BA}} = \dfrac{{kQQ}}{{{d^2}}} = \dfrac{{k{Q^2}}}{{{d^2}}}\] ………(ii)
force exerted by point D on point A, using coulomb’s law from equation (i) we get,
\[{F_{DA}} = \dfrac{{kQQ}}{{{d^2}}} = \dfrac{{k{Q^2}}}{{{d^2}}}\] …………(iii)
To find the force exerted by point C on point A, we need to find the distance between point A and C that is the length of the diagonal of the square.
We have the formula for length of diagonal of a square as,
\[D = \sqrt 2 a\] where \[a\] is the side of the square.
Here side of the square is \[d\] so, the length of the diagonal will be
\[D = \sqrt 2 d\]
We redraw the diagram using this length of the diagonal.
Now, force exerted by point C on point A using coulomb’s law from equation (i), we get
\[{F_{CA}} = \dfrac{{kQQ}}{{{{\left( {\sqrt 2 d} \right)}^2}}}\]
\[ \Rightarrow {F_{CA}} = \dfrac{{k{Q^2}}}{{2{d^2}}}\] (iv)
Now, we can find the total force exerted on point A by adding the force exerted by the other three points.
Now, we draw the diagram with the forces \[{F_{BA}}\], \[{F_{DA}}\] and \[{F_{CA}}\] which are along BA, DA and CA respectively.
The resultant of forces \[{F_{BA}}\] and \[{F_{DA}}\] will be along CA, the resultant of forces \[{F_{BA}}\] and \[{F_{DA}}\] is
\[F' = \sqrt {{F_{BA}}^2 + {F_{DA}}^2} \]
Putting the values of \[{F_{BA}}\] and \[{F_{DA}}\] we get,
\[F' = \sqrt {{{\left( {\dfrac{{k{Q^2}}}{{{d^2}}}} \right)}^2} + {{\left( {\dfrac{{k{Q^2}}}{{{d^2}}}} \right)}^2}} \]
\[ \Rightarrow F' = \sqrt 2 \dfrac{{k{Q^2}}}{{{d^2}}}\] (v)
The forces \[F'\] and \[{F_{CA}}\] are along same direction, so the net force on point A will be addition of these two forces that is,
\[{F_{net}} = F' + {F_{CA}}\]
Putting the values of \[F'\] and \[{F_{CA}}\], we get
\[{F_{net}} = \sqrt 2 \dfrac{{k{Q^2}}}{{{d^2}}} + \dfrac{{k{Q^2}}}{{2{d^2}}}\]
\[ \Rightarrow {F_{net}} = \left( {\sqrt 2 + \dfrac{1}{2}} \right)\dfrac{{k{Q^2}}}{{{d^2}}}\]
Now, putting the values of \[Q\] , \[d\] and \[k = 9 \times {10^9}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 2}}}}\] we get,
\[{F_{net}} = \left( {\sqrt 2 + \dfrac{1}{2}} \right)\dfrac{{\left( {9 \times {{10}^9}} \right){{\left( {16 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {0.2} \right)}^2}}}\]
\[ \Rightarrow {F_{net}} = \left( {\dfrac{{2\sqrt 2 + 1}}{2}} \right)\dfrac{{9 \times 256 \times {{10}^{ - 3}}}}{{0.04}}\]
\[ \Rightarrow {F_{net}} = \left( {2\sqrt 2 + 1} \right) \times 9 \times 32 \times {10^{ - 1}}\]
\[ \ therefore{F_{net}} = 110.3\,{\text{N}}\]
Therefore, the force on any charge will be \[110.3\,{\text{N}}\].
Note:We have used the coulomb’s law. Coulomb states that force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the centers of the two charges. If the distance between the charges is increased the force between them reduces and if the distance between the charges is reduced the force between them increases.
Complete step by step answer:
Given, charge on four corners of a square is, \[Q = 16\,\mu C = 16 \times {10^{ - 6}}\,{\text{C}}\]
Side of the square, \[d = {\text{0}}{\text{.2}}\,{\text{m}}\]
Here, we will use coulomb’s law, which states force between two charges is given as,
\[F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}\] ……………(i)
where \[{q_1}\] and \[{q_2}\] are the two charges, \[r\] is the distance between the centre of two charges and \[k\] is the proportionality constant.
Let ABCD be the square with side \[d = {\text{0}}{\text{.2}}\,{\text{m}}\] and point charge \[Q = 16\,\mu C\] on each corner of the square.
We draw the diagram with the given information.
We are asked to calculate the force on any of the charges. The force on any of the charges will be the same.
Let us calculate the force on point A. We will now see the force exerted by the other three points on point A.Force exerted by point B on point A, using coulomb’s law from equation (i) we get,
\[{F_{BA}} = \dfrac{{kQQ}}{{{d^2}}} = \dfrac{{k{Q^2}}}{{{d^2}}}\] ………(ii)
force exerted by point D on point A, using coulomb’s law from equation (i) we get,
\[{F_{DA}} = \dfrac{{kQQ}}{{{d^2}}} = \dfrac{{k{Q^2}}}{{{d^2}}}\] …………(iii)
To find the force exerted by point C on point A, we need to find the distance between point A and C that is the length of the diagonal of the square.
We have the formula for length of diagonal of a square as,
\[D = \sqrt 2 a\] where \[a\] is the side of the square.
Here side of the square is \[d\] so, the length of the diagonal will be
\[D = \sqrt 2 d\]
We redraw the diagram using this length of the diagonal.
Now, force exerted by point C on point A using coulomb’s law from equation (i), we get
\[{F_{CA}} = \dfrac{{kQQ}}{{{{\left( {\sqrt 2 d} \right)}^2}}}\]
\[ \Rightarrow {F_{CA}} = \dfrac{{k{Q^2}}}{{2{d^2}}}\] (iv)
Now, we can find the total force exerted on point A by adding the force exerted by the other three points.
Now, we draw the diagram with the forces \[{F_{BA}}\], \[{F_{DA}}\] and \[{F_{CA}}\] which are along BA, DA and CA respectively.
The resultant of forces \[{F_{BA}}\] and \[{F_{DA}}\] will be along CA, the resultant of forces \[{F_{BA}}\] and \[{F_{DA}}\] is
\[F' = \sqrt {{F_{BA}}^2 + {F_{DA}}^2} \]
Putting the values of \[{F_{BA}}\] and \[{F_{DA}}\] we get,
\[F' = \sqrt {{{\left( {\dfrac{{k{Q^2}}}{{{d^2}}}} \right)}^2} + {{\left( {\dfrac{{k{Q^2}}}{{{d^2}}}} \right)}^2}} \]
\[ \Rightarrow F' = \sqrt 2 \dfrac{{k{Q^2}}}{{{d^2}}}\] (v)
The forces \[F'\] and \[{F_{CA}}\] are along same direction, so the net force on point A will be addition of these two forces that is,
\[{F_{net}} = F' + {F_{CA}}\]
Putting the values of \[F'\] and \[{F_{CA}}\], we get
\[{F_{net}} = \sqrt 2 \dfrac{{k{Q^2}}}{{{d^2}}} + \dfrac{{k{Q^2}}}{{2{d^2}}}\]
\[ \Rightarrow {F_{net}} = \left( {\sqrt 2 + \dfrac{1}{2}} \right)\dfrac{{k{Q^2}}}{{{d^2}}}\]
Now, putting the values of \[Q\] , \[d\] and \[k = 9 \times {10^9}\,{\text{N}}{{\text{m}}^{\text{2}}}{{\text{C}}^{{\text{ - 2}}}}\] we get,
\[{F_{net}} = \left( {\sqrt 2 + \dfrac{1}{2}} \right)\dfrac{{\left( {9 \times {{10}^9}} \right){{\left( {16 \times {{10}^{ - 6}}} \right)}^2}}}{{{{\left( {0.2} \right)}^2}}}\]
\[ \Rightarrow {F_{net}} = \left( {\dfrac{{2\sqrt 2 + 1}}{2}} \right)\dfrac{{9 \times 256 \times {{10}^{ - 3}}}}{{0.04}}\]
\[ \Rightarrow {F_{net}} = \left( {2\sqrt 2 + 1} \right) \times 9 \times 32 \times {10^{ - 1}}\]
\[ \ therefore{F_{net}} = 110.3\,{\text{N}}\]
Therefore, the force on any charge will be \[110.3\,{\text{N}}\].
Note:We have used the coulomb’s law. Coulomb states that force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between the centers of the two charges. If the distance between the charges is increased the force between them reduces and if the distance between the charges is reduced the force between them increases.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

Why should a magnesium ribbon be cleaned before burning class 12 chemistry CBSE

