Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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

Grade:12

3 Answers

Amritpal Singh Saini
134 Points
2 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
2 years ago
Dear Amritpal Singh,
 
Please stop asking for email ID and contact number of questioners as it is against forum policy. You will not be rewarded for these type of answers and your account may get blocked if you do not stop this. Please check the guidelines here – Forum point Policy – https://www.askiitians.com/forum-point-policy/
 
Thanks
Forum Team
Vivek Singh
13 Points
one month 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
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free