Question

Suppose that the magnitude of the force between two charged particles that are 2 cm apart is 50 n. What will the distance r between the particles be when the magnitude of the force between them is 200 n?

Answer 1

The distance r between the particles will be 1 cm.

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Further explanation

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

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There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

$\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }$

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

Let's tackle the problem now !

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Given:

initial distance = d₁ = 2 cm

initial force = F₁ = 50 N

final force = F₂ = 200 N

Unknown:

final distance = d₂ = ?

Solution:

$F_1 : F_2 = k \frac{q_1 q_2}{(d_1)^2} : k \frac{q_1 q_2}{(d_2)^2}$

$F_1 : F_2 = \frac{1}{(d_1)^2} : \frac{1}{(d_2)^2}$

$F_1 : F_2 = (d_2)^2 : (d_1)^2$

$50 : 200 = (d_2)^2 : (2)^2$

$1 : 4 = (d_2)^2 : 4$

$(d_2)^2 = 4 \div 4$

$d_2 = 1 \texttt{ cm}$

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Learn more

• The three resistors : brainly.com/question/9503202
• A series circuit : brainly.com/question/1518810
• Compare and contrast a series and parallel circuit : brainly.com/question/539204

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Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

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Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces

Answer 2

Answer:

The distance between the particles have to be

x=0.01m or 1 cm

Explanation:

Using the law of Coulomb for two charge particles

$F=K*\frac{q1*q2}{r^{2} }$

Notice you don't know the value of K or q1 or q2 however the value is the same in both cases, so you don't need to know exactly the value of each one you can put constant in the equation like this:

$F=K*\frac{q1*q2}{r^{2} }$

$50N= K*q1*q2*\frac{1}{(0.02m)^{2} } \\k*q1*q2=K$

$K= 50N*(0.02m)^{2}\\ K=0.02$ C

Now to find r can use the value of K

$F=K*\frac{1}{r^{2} } \\r^{2}=\frac{K}{F} \\ r=\sqrt{\frac{0.02 C}{200N} } \\r=0.01m$