Two charges of $7 C$ $7 C$ and $5 C$ $5 C$ are positioned on a line at points $2$ $2$ and $- 4$ $-4$ , respectively. What is the net force on a charge of $- 4 C$ $-4 C$ at $0$ $0$ ?

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Two charges of $7 C$ $7 C$ and $5 C$ $5 C$ are positioned on a line at points $2$ $2$ and $- 4$ $-4$ , respectively. What is the net force on a charge of $- 4 C$ $-4 C$ at $0$ $0$ ?

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Answer 1 · 2016-03-06T02:56:58

$= 51.75 \cdot 10^{9} N$ $=51.75*10^9N$

Explanation:

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By Coulomb's law we Know that force between two charges Q and q situated at a distance r is given by the formula $F = k \frac{q Q}{r^{2}}$ $F=k(qQ)/r^2$ ,where k =Coulomb's constant= $9 \cdot 10^{9} N m^{2} C^{- 2}$ $9*10^9Nm^2C^-2$

Applying this formula the attractive force of 7C on -4C
= $9 \cdot 10^{9} \cdot 4 \cdot \frac{7}{2^{2}} N = 63 \cdot 10^{9} N$ $9*10^9*4*7/2^2N=63*10^9N$
The attractive force of 5C on -4C = $9 \cdot 10^{9} \cdot 4 \cdot \frac{5}{4^{2}} N = 11.25 \cdot 10^{9} N$ $9*10^9*4*5/4^2N=11.25*10^9N$

These two forces are oppositely directed so the net force will be towards right and its magnitude $= (63 - 11.25) \cdot 10^{9} N = 51.75 \cdot 10^{9} N$ $=(63-11.25)*10^9N=51.75*10^9N$

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Two charges of $7 C$ $7 C$ and $5 C$ $5 C$ are positioned on a line at points $2$ $2$ and $- 4$ $-4$ , respectively. What is the net force on a charge of $- 4 C$ $-4 C$ at $0$ $0$ ?

1 Answer

Explanation:

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Two charges of 7C 7 C and 5C 5 C are positioned on a line at points 2 2 and −4 -4 , respectively. What is the net force on a charge of −4C -4 C at 0 0 ?

1 Answer

Explanation:

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Two charges of $7 C$ $7 C$ and $5 C$ $5 C$ are positioned on a line at points $2$ $2$ and $- 4$ $-4$ , respectively. What is the net force on a charge of $- 4 C$ $-4 C$ at $0$ $0$ ?