Physics, 30.07.2019 22:30 andrewmena05

A+1.0 c charged object and a +2.0 c charged object are separated by 100 m. where should a -1.0x10-3 c charged object be placed on a line between the positively charged objects so that the net electrical force exerted on the negatively charged object is zero?

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Answer from: ronaldotheexplorer12

The negative charge should be placed 41 m from +1.0 C charged object

Further explanation

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

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 !

Given:

Q₁ = +1.0 C

Q₂ = +2.0 C

R₁ = x

R₂ = 100 - x

q = -1.0 × 10⁻³ C

Unknown:

x = ?

Solution:

In order to produce a total force = 0, the electric forces acting on this negative charge must be in the opposite direction and equal in magnitude.

We assume the distance of this negative charge to the positive charge +1.0 C is x, then:

F_1 = F_2

$k\frac{Q_1q}{R_1^2} = k\frac{Q_2q}{R_2^2}$

$\frac{Q_1}{R_1^2} = \frac{Q_2}{R_2^2}$

$\frac{1}{x^2} = \frac{2}{(100 - x)^2}$

$\sqrt{\frac{1}{x^2}} = \sqrt{\frac{2}{(100 - x)^2}}$

$\frac{1}{x} = \frac{\sqrt{2}}{100 - x}$

$100 - x = \sqrt{2}x$

$x + \sqrt{2}x = 100$

$x ( 1 + \sqrt{2} ) = 100$

$x = \frac{100}{1 + \sqrt{2}}$

$x = 100(\sqrt{2} - 1) ~ m$

$x \approx 41 ~ m$

Learn moreThe three resistors : A series circuit : Compare and contrast a series and parallel circuit : Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

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

A+1.0 c charged object and a +2.0 c charged object are separated by 100 m. where should a -1.0x10-3

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

answer would be letter choice (a), a ball resting on a shelf. would be another example of potential energy > stored energy > not moving.

a ball spinning on a finger > kinetic energy > moving energy

a ball rolling on the ground > kinetic energy > moving energy

a ball flying through the air > kinetic energy > moving energy

hope that : )

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