Two identical conducting spheres, fixed in space, attract each other with an electrostatic force of $ 0.108N $ when separated by $ 50.0cm $ , centre - to – centre. A thin conducting wire then connects the spheres, when the wire is removed, the spheres repel each other with an electrostatic force of $ 0.0360N $ . What were the initial charges on the spheres?
Answer
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Hint: To solve this question we have to know about force repulsion, we know that repulsion is a power between at least two comparable or like charges. Two charges of comparable qualities pull away from one another. We have to also know that, a repulsive force or power of a speeding up universe, which as indicated by specific speculations makes planets and matter get farther and farther separated. Like charges repulsing as per Coulomb's law. Frightful power (attraction) between magnets of inverse direction.
Complete answer:
Let us consider the initial charges are equal to $ {q_1} $ and $ - {q_2} $
Now, according to the question, the force of attraction is $ 0.108N $ which is equal to $ = \dfrac{1}{{4\pi \varepsilon }}\dfrac{{{q_1}{q_2}}}{{{{0.5}^2}}} $
Now, the charges after the connection is made and removed $ = \dfrac{{({q_1} - {q_2})}}{2} $ and $ = \dfrac{{({q_1} - {q_2})}}{2} $
We can write the force repulsion is equal to $ 0.0360N $ which is equal to, $ = \dfrac{{{{({q_1} - {q_2})}^2}}}{{4 \times {{0.5}^2}}}\dfrac{1}{{4\pi \varepsilon }} $
Therefore,
$ {q_1}{q_2} = 3.0012 \times {10^{ - 12}} $
And, $ {({q_1} - {q_2})^2} = 4 \times {10^{ - 12}} $
Therefore, $ {q_2} = {q_1} = 1\mu F $ .
Note:
We should also know about the charges. What is charge and what are the functions of the charges. We can say, In physics, charge, otherwise called electric charge, electrical charge, or electrostatic charge and represented q, is a quality of a unit of issue that communicates the degree to which it has more or less electrons than protons. In particles, the electron conveys a negative rudimentary or unit charge; the proton conveys a positive charge.
Complete answer:
Let us consider the initial charges are equal to $ {q_1} $ and $ - {q_2} $
Now, according to the question, the force of attraction is $ 0.108N $ which is equal to $ = \dfrac{1}{{4\pi \varepsilon }}\dfrac{{{q_1}{q_2}}}{{{{0.5}^2}}} $
Now, the charges after the connection is made and removed $ = \dfrac{{({q_1} - {q_2})}}{2} $ and $ = \dfrac{{({q_1} - {q_2})}}{2} $
We can write the force repulsion is equal to $ 0.0360N $ which is equal to, $ = \dfrac{{{{({q_1} - {q_2})}^2}}}{{4 \times {{0.5}^2}}}\dfrac{1}{{4\pi \varepsilon }} $
Therefore,
$ {q_1}{q_2} = 3.0012 \times {10^{ - 12}} $
And, $ {({q_1} - {q_2})^2} = 4 \times {10^{ - 12}} $
Therefore, $ {q_2} = {q_1} = 1\mu F $ .
Note:
We should also know about the charges. What is charge and what are the functions of the charges. We can say, In physics, charge, otherwise called electric charge, electrical charge, or electrostatic charge and represented q, is a quality of a unit of issue that communicates the degree to which it has more or less electrons than protons. In particles, the electron conveys a negative rudimentary or unit charge; the proton conveys a positive charge.
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