Question

Force acts on charge 5uC due to charge 10uC is F1, and force acts on charge 10uC due to charge 5uC is F2 for same separation. Ratio of F1 and F2 is
a) 1:2 b) 2:1 c) 1:1
d) 1:5​

Answer 1

Answer :

• The ratio of charge of 5μC acts on the charge of 10μC and when the charge of 10μC acts on 5μC is 1 : 1.

Explanation :

Given :

• Charge, C1 = 5μC
• Charge, C2 = 10μC
• Force when charge of 5μC acts on the charge of 10μC, F = F1.
• Force when charge of 10μC acts on the charge of 5μC, F = F2.
• Distance of seperation is same in both the cases.

To find :

• The ratio of the forces F1 and F2 = ?

Knowledge required :

Formula for force between two charges :

$\boxed{\therefore \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Qq}{r^{2}}}}$

Where,

• $\sf{\dfrac{1}{4\pi\varepsilon_{0}}$ = Constant
• Q & q = Charges
• r = Distance of seperation

Solution :

According to the question, the seperation between the charges are same, when the charge of 5μC acts on the charge of 10μC and when the charge of 10μC acts on 5μC, so we can write them as a single variable.

Let the distance of seperation between the charges be r.

First let us find the force acting on the charge of 10μc by the charge of 5μC.

By using the formula for electric force and substituting the values in it, we get :

$:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Qq}{r^{2}}}$

$:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{5 \times 10}{r^{2}}}$

$:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}$

$\boxed{\therefore \sf{F_{1} = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}}$

Hence the force acting on 10μC by the force of 5μC is $\sf{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}$

Now let us find the force acting on the charge of 5μc by the charge of 10μC.

By using the formula for electric force and substituting the values in it, we get :

$:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Qq}{r^{2}}}$

$:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{10 \times 5}{r^{2}}}$

$:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}$

$\boxed{\therefore \sf{F_{2} = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}}$

Hence the force acting on 10μC by the force of 5μC is $\sf{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}$

Now to find the ratio of the forces acting on both the charges. i.e, F1/F2.

$:\implies \sf{\dfrac{F_{1}}{F_{2}} = \dfrac{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}}$

$:\implies \sf{\dfrac{F_{1}}{F_{2}} = \dfrac{1}{\cancel{4\pi\varepsilon_{0}}}\cancel{\dfrac{50}{r^{2}}} \times \cancel{4\pi\varepsilon_{0}}\cancel{\dfrac{r^{2}}{50}}}$

$:\implies \sf{\dfrac{F_{1}}{F_{2}} = \dfrac{1}{1}}$

$\boxed{\therefore \sf{F_{1} : F_{2} = 1 : 1}}$

Therefore,

• The ratio of the forces F1 and F2 is 1 : 1.

Answer 2

Answer:

Answer :

The ratio of charge of 5μC acts on the charge of 10μC and when the charge of 10μC acts on 5μC is 1 : 1.

Explanation :

Given :

Charge, C1 = 5μC

Charge, C2 = 10μC

Force when charge of 5μC acts on the charge of 10μC, F = F1.

Force when charge of 10μC acts on the charge of 5μC, F = F2.

Distance of seperation is same in both the cases.

To find :

The ratio of the forces F1 and F2 = ?

Knowledge required :

Formula for force between two charges :

\begin{gathered}\boxed{\therefore \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Qq}{r^{2}}}} \\ \\ \end{gathered}

∴F=

4πε

0

1

r

2

Qq

Where,

\sf{\dfrac{1}{4\pi\varepsilon_{0}} = Constant

Q & q = Charges

r = Distance of seperation

Solution :

According to the question, the seperation between the charges are same, when the charge of 5μC acts on the charge of 10μC and when the charge of 10μC acts on 5μC, so we can write them as a single variable.

Let the distance of seperation between the charges be r.

First let us find the force acting on the charge of 10μc by the charge of 5μC.

By using the formula for electric force and substituting the values in it, we get :

\begin{gathered}:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Qq}{r^{2}}} \\ \\ \end{gathered}

:⟹F=

4πε

0

1

r

2

Qq

\begin{gathered}:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{5 \times 10}{r^{2}}} \\ \\ \end{gathered}

:⟹F=

4πε

0

1

r

2

5×10

\begin{gathered}:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}} \\ \\ \end{gathered}

:⟹F=

4πε

0

1

r

2

50

\begin{gathered}\boxed{\therefore \sf{F_{1} = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}} \\ \\ \end{gathered}

∴F

1

=

4πε

0

1

r

2

50

Hence the force acting on 10μC by the force of 5μC is \begin{gathered}\sf{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}} \\ \\ \end{gathered}

4πε

0

1

r

2

50

Now let us find the force acting on the charge of 5μc by the charge of 10μC.

By using the formula for electric force and substituting the values in it, we get :

\begin{gathered}:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{Qq}{r^{2}}} \\ \\ \end{gathered}

:⟹F=

4πε

0

1

r

2

Qq

\begin{gathered}:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{10 \times 5}{r^{2}}} \\ \\ \end{gathered}

:⟹F=

4πε

0

1

r

2

10×5

\begin{gathered}:\implies \sf{F = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}} \\ \\ \end{gathered}

:⟹F=

4πε

0

1

r

2

50

\begin{gathered}\boxed{\therefore \sf{F_{2} = \dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}} \\ \\ \end{gathered}

∴F

2

=

4πε

0

1

r

2

50

Hence the force acting on 10μC by the force of 5μC is \begin{gathered}\sf{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}} \\ \\ \end{gathered}

4πε

0

1

r

2

50

Now to find the ratio of the forces acting on both the charges. i.e, F1/F2.

\begin{gathered}:\implies \sf{\dfrac{F_{1}}{F_{2}} = \dfrac{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}{\dfrac{1}{4\pi\varepsilon_{0}}\dfrac{50}{r^{2}}}} \\ \\ \end{gathered}

:⟹

F

2

F

1

=

4πε

0

1

r

2

50

4πε

0

1

r

2

50

\begin{gathered}:\implies \sf{\dfrac{F_{1}}{F_{2}} = \dfrac{1}{\cancel{4\pi\varepsilon_{0}}}\cancel{\dfrac{50}{r^{2}}} \times \cancel{4\pi\varepsilon_{0}}\cancel{\dfrac{r^{2}}{50}}} \\ \\ \end{gathered}

:⟹

F

2

F

1

=

4πε

0

1

r

2

50

×

4πε

0

50

r

2

\begin{gathered}:\implies \sf{\dfrac{F_{1}}{F_{2}} = \dfrac{1}{1}} \\ \\ \end{gathered}

:⟹

F

2

F

1

=

1

1

\begin{gathered}\boxed{\therefore \sf{F_{1} : F_{2} = 1 : 1}} \\ \\ \end{gathered}

∴F

1

:F

2

=1:1

Therefore,

The ratio of the forces F1 and F2 is 1 : 1.