Question

(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10−7 C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer 1

(a) charge on each copper sphere, q = 6.5 × 10^-7 C.

seperation between spheres , r = 50cm = 0.5m

from Coulomb's force, F = $\frac{kq_1q_2}{r^2}$

here, $q_1=q_2=q$

= 9 × 10^9 × (6.5 × 10^-7)²/(0.5)²

= 1.52 × 10^-2 N

hence, mutual force of electrostatic repulsion , F = 1.52 × 10^-2 N

(b) if each sphere is charged doubled the above amount and distance between them is halved.

i.e., $q'_1=2q$ , $q'_2=2q$ and $r'=\frac{r}{2}$

so, F' = $\frac{kq'_1q'_2}{r'^2}$

= $\frac{k(2q)(2q)}{\left(\frac{r}{2}\right)^2}$

= 16 × kq²/r²

= 16 × F

= 16 × 1.52

= 24.32 × 10^-2 N or 0.24 N

Answer 2

Explanation:



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PHYSICS

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7 C? The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

December 26, 2019Meghana Dungriwala

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VIDEO EXPLANATION



ANSWER

(a) 

Charge on sphere A, qA=6.5×10−7C 

Charge on sphere B, qB=6.5×10−7C

Distance between the spheres, r=50cm=0.5m

Force of repulsion between the two spheres,

F=4π∈0r2qAqB

Where, ∈0= Free space permittivity

4π∈01=9×109Nm2C2

∴F=(0.5)29×109×(6.5×10−7)2

         =1.52×10−2N

Therefore, the force between the two spheres is 1.52×10