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Byju's Answer

Standard XIII

Physics

Coulomb's Law - Mathematical Form

Two small sph...

Question

Two small spherical balls each carrying a charge Q = 10 μC (10 micro-coulomb) are suspended by two insulating threads of equal lengths 1m each, from a point fixed in the ceiling. It is found that in equilibrium threads are separated by an angle 60^o c between them, as shown in the figure. What is the tension in the threads
(Given : $\frac{1}{(4 π ϵ_{0})} = 9 \times 10^{9} N M^{2} / c^{2}$ )

18 N

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1.8 N

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0.18 N

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None of the above

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Solution

The correct option is B 1.8 N
In the following figure, in equalibrium $F_{e} = T s i n 30^{\circ}, r = 1 m$

$\Rightarrow 9 \times 10^{9} \frac{Q^{2}}{r^{2}} = T \times \frac{1}{2}$
$\Rightarrow 9 \times 10^{9} . \frac{(10 \times 10^{- 6})^{2}}{1^{2}} = T \times \frac{1}{2} \Rightarrow T = 1.8 N$

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Two small identical balls P and Q, each of mass $\frac{\sqrt{3}}{10}$ gram, carry identical charges and are suspended by threads of equal lengths. At equilibrium, they position themselves as shown in Fig. 20.5 what is the charge on each ball. Given $\frac{1}{4 π ε_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}$ and take $g = 10 N / K g .$

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The correct option is B 1.8 NIn the following figure, in equalibrium Fe=T sin 30∘,r=1m ⇒9×109Q2r2=T×12 ⇒ 9×109.(10×10−6)212=T×12⇒T=1.8 N

The correct option is B 1.8 N
In the following figure, in equalibrium $F_{e} = T s i n 30^{\circ}, r = 1 m$

$\Rightarrow 9 \times 10^{9} \frac{Q^{2}}{r^{2}} = T \times \frac{1}{2}$
$\Rightarrow 9 \times 10^{9} . \frac{(10 \times 10^{- 6})^{2}}{1^{2}} = T \times \frac{1}{2} \Rightarrow T = 1.8 N$