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There are two sphere of radii R and 2R having charges Q and 2Q respectively and separated far away. these two sphere are connected with a cell of emf v voltts now through a switch S. when the switch is sed , find the final charge on the smaller sphere. \lbrack v=KQ/2R\rbrack?
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There are two sphere of radii R and 2R having charges Q and 2Q respect...
Introduction
We are given two spheres with radii R and 2R, having charges Q and 2Q respectively. These spheres are separated far away from each other and are connected through a cell of electromotive force (emf) v volts. We need to find the final charge on the smaller sphere.
Explanation
To solve this problem, we can follow the concept of charge conservation and the principle of potential difference.
Step 1: Identifying the initial potential difference
Initially, the spheres are separated and not connected. Therefore, the potential difference between them is zero.
Step 2: Connecting the spheres
When the switch S is closed, the spheres are connected through the cell of emf v volts. This causes charges to redistribute between the spheres until an equilibrium is reached.
Step 3: Charge redistribution
As charges redistribute, the potential difference between the spheres changes. Let's denote the final charge on the smaller sphere as q.
Step 4: Applying the principle of potential difference
According to the principle of potential difference, the potential difference across the spheres is given by v = (Kq/2R) - (K(2Q - q)/4R), where K is the electrostatic constant.
Simplifying the equation, we get:
v = (Kq/2R) - (K(2Q - q)/4R)
v = (Kq/2R) - (2KQ - Kq)/4R
v = (Kq - 4KQ + Kq)/4R
v = (2Kq - 4KQ)/4R
v = (Kq - 2KQ)/2R
Step 5: Finding the final charge on the smaller sphere
To find the final charge on the smaller sphere, we need to solve the equation obtained in step 4 for q. Rearranging the equation, we get:
(Kq - 2KQ) = 2Rv
Kq = 2Rv + 2KQ
q = (2Rv + 2KQ)/K
Therefore, the final charge on the smaller sphere is (2Rv + 2KQ)/K.
Conclusion
The final charge on the smaller sphere, when the switch is closed, is given by (2Rv + 2KQ)/K. This result is obtained by applying the principle of potential difference and considering charge conservation during the redistribution of charges between the spheres.
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There are two sphere of radii R and 2R having charges Q and 2Q respectively and separated far away. these two sphere are connected with a cell of emf v voltts now through a switch S. when the switch is sed , find the final charge on the smaller sphere. \lbrack v=KQ/2R\rbrack?
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