# A clock face has positive charges +1C,+2C,+3C.........+12C placed at corresponding every hour numbers of a Find net force at the center O of the clock

Amritpal Singh Saini
134 Points
5 years ago
Is there any charge at the centre ?? I think so I have come across such a same question many times and in the centre there is always 1C charge!
Forum Team
94 Points
5 years ago
Dear Amritpal Singh,

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Forum Team
Vivek Singh
13 Points
3 years ago
Let us consider a 1C charge at the centre O of the clock which is at the distance of 1m from all the hour numbers.
Now , force on O due to 1C charge is
F = KQ1 Q2 / r^2 = K× 1×1 / (1)^2
F = K
Now , force on O due to 2C charge is
F = KQ1 Q2 / r^2 = K×1×2 / (1)^2
F=2K
Now , force on O due to 3C charge is
F = KQ1 Q2 / r^2 = K×1×3 / (1)^2
F = 3K
Now force on O due to 4C charge is
F = KQ1 Q2 / r^2 = K×1×4/(1)^
F= 4K
Now force on O due to 5C charge is
F = KQ1 Q2 / r^2 = K×1×5/(1)^2
F = 5K
Now force on O due to 6C charge is
F = KQ1 Q2 / r^2 = K×1×6/(1)^2
F= 6K
Now force on O due to 7C charge is
F= KQ1 Q2 /r^2 = K×1×7 / (1)^2
F= 7K
Now force on O due to 8C charge is
F= KQ1 Q2 / r^2 = K×1×8 /(1)^2
F= 8K
Now force on O due to 9C charge is
F= KQ1 Q2 / r^2 = K×1×9 /(1)^2
F= 9K
Now force on O due to 10C charge is
F=KQ1 Q2/ r^2 = K×1×10 /(1)^2
F= 10K
Now force on O due to 11C charge is
F = KQ1 Q2 / r^2 = K×1×11 /(1)^2
F= 11K
Now force on O due to 12C charge is
F=KQ1 Q2 /r^2 = K×1×12 /(1)^2
F= 12K
As the force due to 1C and 7C is in opposite direction. So the net force due to them is = 7K - 1K = 6K
Similarly force due to 2C and 8C is in opposite direction.  So the net force due to them is = 8K -- 2K = 6K
Similarly force due to 3C and 9C is in opposite direction.  So the net force due to them is
= 9K-3K = 6K
Similarly force due to 10C and 4C is in opposite direction.  So net force due to them is
= 10K - 4K = 6k
Similarly force due to 11C and 5C is in opposite direction.  So net force due to them is
= 11K - 5k = 6K
Similarly force due to 12C and 6C is in opposite direction. So net force due to them is
= 12K - 6K = 6K
It means that force = 6K is present all around the charge .
Now resolving the forces,
F net = (6K + 6K Cos30 + 6KCos60 + 6K Cos30 + 6K Cos60 ) i + (-6K) j
= (6K + 12Kcos30 + 12Kcos60) i - 6K j
=(6K + 12K × under root 3 / 2 + 12 K × 1/2)i - 6K j
=(6K + 6under root 3 K + 6K) i - 6K j
= (12K + 6under root 3 K) i - 6K j
= 6K ( 2+ under root 3 ) i - 6K j