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Two charges of # 9 C # and # -8 C# are positioned on a line at points # -3 # and # -1 #, respectively. What is the net force on a charge of # 7 C# at # 0 #?

Answer 1

Olivia Askew

#441×10^9 N#

Here,the #9C# charge tries to push the #7C# charge towards positive direction.

And the value of that force #F1# is #(9×10^9)×9×7/(3)^2 N# i.e #63×10^9 N#

Whereas, the #-8C# charge tries to pull the #7C# charge towards its position i.e in negative direction.

And the amount of force #F2# is #(9×10^9)×(-8)×7/(-1)^2 N# or,#504×10^9 N#

So,net force acting is towards negative direction and it is #F2 - F1# i.e #(504-63)×10^9 N# or, #441×10^9 N#

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Answer 2

Parker Abbott

( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} ), where ( k ) is Coulomb's constant (( 8.99 \times 10^9 , \text{N m}^2/\text{C}^2 )), ( q_1 ) and ( q_2 ) are the charges, and ( r ) is the separation distance. The net force is the vector sum of individual forces.

( F_{\text{net}} = F_{01} + F_{02} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.