JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 15)
Ten charges are placed on the circumference
of a circle of radius R with constant angular
separation between successive charges.
Alternate charges 1, 3, 5, 7, 9 have charge (+q)
each, while 2, 4, 6, 8, 10 have charge (–q) each.
The potential V and the electric field E at the
centre of the circle are respectively.
(Take V = 0 at infinity)
(Take V = 0 at infinity)
V = 0; E = 0
$$V = {{10q} \over {4\pi {\varepsilon _0}R}}$$; $$E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}$$
$$V = {{10q} \over {4\pi {\varepsilon _0}R}}$$; E = 0
V = 0; $$E = {{10q} \over {4\pi {\varepsilon _0}{R^2}}}$$
Explanation
Net charge = 5q - 5q = 0
Potential of centre = V = $${{K\sum q } \over r}$$
VC = $${{K\left( 0 \right)} \over r}$$ = 0_5th_September_Evening_Slot_en_15_1.png)
Let E be electric field produced by each charge at the centre, then resultant electric field will be
_5th_September_Evening_Slot_en_15_2.png)
EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.( From symmetric property of vector)
Potential of centre = V = $${{K\sum q } \over r}$$
VC = $${{K\left( 0 \right)} \over r}$$ = 0
Let E be electric field produced by each charge at the centre, then resultant electric field will be
EC = 0, Since equal electric field vectors are acting at equal angle so their resultant is equal to zero.( From symmetric property of vector)
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