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A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm in allowed to touch it for some time. If both the spheres are separated, then surface density of charges on the spheres will be in the ratio of
- a)1:4
- b)4:1
- c)1:2
- d)2:!
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
A conducting sphere of radius 10 cm is charged with 10 μC. Another ...
Given data:
- Radius of first sphere (charged sphere), \(r_1 = 10 \, \text{cm}\)
- Charge on first sphere, \(Q_1 = 10 \, \mu C\)
- Radius of second sphere (uncharged sphere), \(r_2 = 20 \, \text{cm}\)
Step 1: Determine the charge on the uncharged sphere after touching
- When the uncharged sphere touches the charged sphere, charge will transfer from the charged sphere to the uncharged sphere until they reach the same potential.
- Using conservation of charge, the final total charge is equal to the initial total charge.
- Let \(Q_2\) be the final charge on the uncharged sphere.
- Therefore, \(Q_1 + Q_2 = Q_1' + Q_2'\)
- \(Q_2 = Q_1' + Q_2' - Q_1 = Q_2'\)
- The final charges on both spheres will be equal.
Step 2: Calculate the surface charge density ratio
- Surface charge density, \(\sigma = \frac{Q}{4\pi r^2}\)
- For the first sphere, \(\sigma_1 = \frac{Q_1'}{4\pi r_1^2}\)
- For the second sphere, \(\sigma_2 = \frac{Q_2'}{4\pi r_2^2}\)
- From step 1, \(Q_1' = Q_2'\)
- Therefore, \(\frac{\sigma_1}{\sigma_2} = \frac{Q_1'}{4\pi r_1^2} \times \frac{4\pi r_2^2}{Q_2'} = \frac{Q_1'}{Q_2'} = 1\)
- Hence, the surface density of charges on the spheres will be in the ratio of 1:1, which is equivalent to 2:2 or 2:1. Thus, option 'D' is correct.
- Radius of first sphere (charged sphere), \(r_1 = 10 \, \text{cm}\)
- Charge on first sphere, \(Q_1 = 10 \, \mu C\)
- Radius of second sphere (uncharged sphere), \(r_2 = 20 \, \text{cm}\)
Step 1: Determine the charge on the uncharged sphere after touching
- When the uncharged sphere touches the charged sphere, charge will transfer from the charged sphere to the uncharged sphere until they reach the same potential.
- Using conservation of charge, the final total charge is equal to the initial total charge.
- Let \(Q_2\) be the final charge on the uncharged sphere.
- Therefore, \(Q_1 + Q_2 = Q_1' + Q_2'\)
- \(Q_2 = Q_1' + Q_2' - Q_1 = Q_2'\)
- The final charges on both spheres will be equal.
Step 2: Calculate the surface charge density ratio
- Surface charge density, \(\sigma = \frac{Q}{4\pi r^2}\)
- For the first sphere, \(\sigma_1 = \frac{Q_1'}{4\pi r_1^2}\)
- For the second sphere, \(\sigma_2 = \frac{Q_2'}{4\pi r_2^2}\)
- From step 1, \(Q_1' = Q_2'\)
- Therefore, \(\frac{\sigma_1}{\sigma_2} = \frac{Q_1'}{4\pi r_1^2} \times \frac{4\pi r_2^2}{Q_2'} = \frac{Q_1'}{Q_2'} = 1\)
- Hence, the surface density of charges on the spheres will be in the ratio of 1:1, which is equivalent to 2:2 or 2:1. Thus, option 'D' is correct.
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A conducting sphere of radius 10 cm is charged with 10 μC. Another uncharged sphere of radius 20 cm in allowed to touch it for some time. If both the spheres are separated, then surface density of charges on the spheres will be in the ratio ofa)1:4b)4:1c)1:2d)2:!Correct answer is option 'D'. Can you explain this answer?
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