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Tju [1.3M]
3 years ago
13

When two point charges are a distance d apart, the magnitude of the electrostatic force between them is F. If the distance betwe

en the point charges is increased to 3d, the magnitude of the electrostatic force between the two charges will be
Physics
2 answers:
dedylja [7]3 years ago
8 0
Decreased by a factor of 9
EastWind [94]3 years ago
3 0

Answer:

The magnitude of the electrostatic force decreases by a factor 9

Explanation:

The electrostatic force between two charges is given by:

F=k\frac{q_1 q_2}{d^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

d is the distance between the two charges

We see that the magnitude of the force is inversely proportional to the square of the distance. If the distance is increased to 3d: d' = 3d, the new electrostatic force would be:

F'=k\frac{q_1 q_2}{(d')^2}=k\frac{q_1 q_2}{(3d)^2}=\frac{1}{9} k\frac{q_1 q_2}{d^2}=\frac{F}{9}

So, the electrostatic force decreases by a factor 9.

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for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

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there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

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s = ut + 1/2att

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The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

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section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

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