Multiple charges and coulombs law

In summary, the problem involves two negative charges of -1.0 X 10^-6 C separated by 0.10 m and a positive charge of 1.0 x 10^-8 C placed midway between them. The task is to determine the electric field and electric potential at the position of the positive charge and either of the negative charges, as well as the electric force experienced by the positive and negative charges. The relevant equations are E=kq2/Kr2, F=eq1, F=kq1q2/Kr2, W=kq2q1/Kr=Vq1, and V=kq2/Kr, where k is the electric constant for water (80.4). The approach
  • #1
chocoiste
1
0

Homework Statement


Two negative charges of -1.0 X 10^-6 C are placed in water and separated by a distance of 0.10 m. A positive charge of 1.0 x 10^-8 C is placed exactly midway between the two negative charges.
Determine:

a) the electric field( magnitude and direction) and electric potential
i)at the position of the positive charge
ii) either negative charge

b) the electric force( magnitude and direction) experienced
i) by the positive charge
ii) by the negative charge


Homework Equations


E=kq2/Kr2 F=eq1 F=kq1q2/Kr2
K is electric constant pertaining to material( in this case water) which is 80.4

W=kq2q1/Kr=Vq1 V=kq2/Kr



The Attempt at a Solution



I don't know where to start.:S Any help will be appreciated.
 
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  • #2
chocoiste said:

Homework Statement


Two negative charges of -1.0 X 10^-6 C are placed in water and separated by a distance of 0.10 m. A positive charge of 1.0 x 10^-8 C is placed exactly midway between the two negative charges.
Determine:

a) the electric field( magnitude and direction) and electric potential
i)at the position of the positive charge
ii) either negative charge

b) the electric force( magnitude and direction) experienced
i) by the positive charge
ii) by the negative charge



Homework Equations


E=kq2/Kr2 F=eq1 F=kq1q2/Kr2
K is electric constant pertaining to material( in this case water) which is 80.4

W=kq2q1/Kr=Vq1 V=kq2/Kr



The Attempt at a Solution



I don't know where to start.:S Any help will be appreciated.

Hello there;

Why don't you start with drawing a diagram and placing the charges into positions,

after you do that consider the E as a vectorial value which affects the charged with the inverse proportional square of radius.So while doing your calculations take note that the values are vectorial not scalar.

Feel free to write if it still looks complicated;

Cryphonus
 
  • #3


I would start by identifying the key information in the problem and organizing it in a way that makes it easier to solve. In this case, the key information is the values of the charges (-1.0 x 10^-6 C and 1.0 x 10^-8 C), the distance between them (0.10 m), and the electric constant for water (80.4). From there, I would use Coulomb's law to calculate the electric field and electric potential at the position of the positive charge and either negative charge.

To calculate the electric field at the position of the positive charge, I would use the equation E = kq/r^2, where k is the electric constant, q is the magnitude of the charge, and r is the distance from the charge. In this case, the distance is 0.05 m (halfway between the two negative charges), so the electric field would be:

E = (80.4)(1.0 x 10^-8 C)/(0.05 m)^2
E = 1.61 x 10^-4 N/C

This tells us that the electric field at the position of the positive charge is 1.61 x 10^-4 N/C, directed towards the negative charges.

To calculate the electric potential at the position of the positive charge, I would use the equation V = kq/r, where V is the electric potential, k is the electric constant, q is the magnitude of the charge, and r is the distance from the charge. In this case, the distance is 0.05 m, so the electric potential would be:

V = (80.4)(1.0 x 10^-8 C)/(0.05 m)
V = 1.61 x 10^-4 V

This tells us that the electric potential at the position of the positive charge is 1.61 x 10^-4 V.

To calculate the electric force experienced by the positive charge, I would use the equation F = kq1q2/r^2, where F is the electric force, k is the electric constant, q1 and q2 are the magnitudes of the two charges, and r is the distance between them. In this case, the distance is 0.10 m, so the electric force would be:

F = (80.4)(1.0 x 10^-8 C)(1
 

Related to Multiple charges and coulombs law

1. What is the definition of multiple charges?

Multiple charges refer to a system of more than one electric charge, such as two or more protons, electrons, or ions.

2. How does coulomb's law describe the interaction between multiple charges?

Coulomb's law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

3. Can coulomb's law be applied to multiple charges that are not stationary?

Yes, coulomb's law can be applied to multiple charges that are in motion, as long as their velocities are not close to the speed of light.

4. What is the unit used to measure electric charge?

The unit of electric charge is the coulomb (C), named after French physicist Charles-Augustin de Coulomb. It is equivalent to approximately 6.24 x 10^18 elementary charges.

5. How does the presence of multiple charges affect the strength of an electric field?

The presence of multiple charges increases the strength of an electric field due to the superposition principle, which states that the total electric field at a point is the vector sum of the individual fields created by each charge.

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