Question

Two identical balls, each with a charge of 2.00 × 10−7 C and a mass of 100 g, are suspended from a common point by two insulating strings, each 50 cm long. The balls are held 5.0 cm apart and then released. Find (a) the electric force on one of the charged balls (b) the components of the resultant force on it along and perpendicular to the string (c) the tension in the string (d) the acceleration of one of the balls. Answers are to be obtained only for the instant just after the release.

Answer 1

Explanation:

Given data in the question  

$q=2.0 \times 10^{-7} \mathrm{C}$

$r=5 \times 10^{-2} \mathrm{m}$

$l=50 \mathrm{cm}=50 \times 10^{-2} \mathrm{m}$

m = 100 g = 0.1 kg

q is Magnitude charges  

r is Separation between the charges

l is string length  

m is ball’s mass

(a) Electric force by Coulomb's Law,            

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q^{2}}{r^{2}}$    

$F=9 \times 10^{9} \times \frac{\left(2.0 \times 10^{-7}\right)^{2}}{\left(5 \times 10^{-2}\right)^{2}}$

$F=9 \times 10^{9} \times \frac{4 \times 10^{-14}}{25 \times 10^{-4}}$

$F=9 \times 10^{9} \times 0.16 \times 10^{-10}$

$F=1.44 \times 10^{-1}$

$F=0.144 N$

(b) The components of the resulting force along it are zero due to mg Tcosθ equilibrium,

F = Tsinθ

(c) string tension  

Tsinθ = F     ..eq^n  (1)

Tcosθ = mg    ...eq^n (2)

Squaring (1) and (2) equations and combining, we get

$\mathrm{T}^{2} \sin ^{2} \theta+\mathrm{T}^{2} \cos ^{2} \theta=\mathrm{F}^{2} \pm(\mathrm{mg})^{2}$  

$\mathrm{T}^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\mathrm{F}^{2} \pm(\mathrm{mg})^{2}$  

We know that sin^2 θ  + cos^2 θ = 1

so

$\mathrm{T}^{2}=\mathrm{F}^{2}+(\mathrm{mg})^{2}$    

$=(0.144)^{2}+(0.1 \times 9.8)^{2}$

$\mathrm{T}=0.986 \mathrm{N}$

Answer 2

Answer:

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