Finding electric field intensity of two spheres

In summary, the question involves two small spheres with equal and opposite charges suspended by light strings in a uniform electric field. The goal is to determine the electric field intensity that will keep the spheres in equilibrium at a given angle theta. Using the equations Felec = kQ1Q2 / r^2 and Tsing theta - mg = 0, we can calculate r in terms of theta and L. The final equation for electric field intensity is E = [(kQ) / (2L cos (theta))^2] + [mgcos (theta) / Qsin (theta)]. All other quantities mentioned, including m, L, Q, theta, g, and k, are used in this equation.
  • #1
an_mui
47
0
Any help on this question is appreciated.
Two small spheres each of mass m are suspended by light strings of light L. A uniform electric field is applied in the x direction. If the spheres have equal and opposite charges of magnitude Q, determine the electric field intensity that enables the spheres to be in equillibrium at angle theta. Express your answers in terms of m, L, Q, theta, g and k.
------------ this is the diagram .
/|\
/ | \
/ | \
O O ****edit: for some reason the diagram is not
(-) (+) aligned properly when i submit the post
from the diagram I came up with two equations.
Felec + Tcos theta - Fexternal = 0
Tsing theta - mg = 0
Felec = kQ1Q2 / r^2
i don't know what L is for since i don't have it in my equations.
 
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  • #2
Are you sure the charges are equal and opposite?? ..In that case the two spheres would attract each other , instead of being at equilibrium. I think the two charges are same , Q each and would repulse each other, with Coulomb's force .

The 'L' given in the question is useful sice you can express the distance 'r' between the spheres in equilibrium state in terms of 'theta' and 'L' , to find out 'theta' finally!

BJ
 
  • #3
yes i am sure the two charges are equal and opposite.
 
  • #4
do you know r?

then calculate r using L and Theeta
 
  • #5
We are to express the answer in terms of m, L, Q , theta, g and k, so we aren't given any numbers. However, my teacher said the two unknowns in this equation are Tension and the electric field intensity
 
  • #6
r = 2*L*cos(theeta)

Fexternal = QE

solve eliminating quantities not required.
 
  • #7
is E equal to ...

E = (kQ1Q2) / (Q2L cos (theta))^2 + mgcos (theta) / Qsin (theta)?
 
  • #8
Q1 =Q2 =Q
so

E = kQ/ [(2L cos (theta))^2] + [mgcos (theta) / Qsin (theta)]
 
Last edited:
  • #9
Q1 = Q2 = Q
hence

E = [(kQ) / (2L cos (theta))^2] + [mgcos (theta) / Qsin (theta)]
 

1. How are the electric field intensity and distance related for two spheres?

The electric field intensity between two spheres is inversely proportional to the square of the distance between them. This means that as the distance between the two spheres increases, the electric field intensity decreases.

2. What is the formula for calculating the electric field intensity between two spheres?

The formula for calculating the electric field intensity between two spheres is E = kQ/r^2, where E is the electric field intensity, k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), Q is the total charge of the spheres, and r is the distance between the centers of the spheres.

3. How do the charges on the two spheres affect the electric field intensity?

The charges on the two spheres affect the electric field intensity by determining the strength and direction of the electric field. If the charges on the spheres are the same, the electric field will repel them and the intensity will be weaker. If the charges are opposite, the electric field will attract them and the intensity will be stronger.

4. Can the electric field intensity between two spheres be negative?

Yes, the electric field intensity between two spheres can be negative. This means that the electric field is directed in the opposite direction of the positive charge and is considered to be an attractive force.

5. What factors can affect the accuracy of finding the electric field intensity between two spheres?

The accuracy of finding the electric field intensity between two spheres can be affected by factors such as the precision of measurements, the distribution of charge on the spheres, and external influences such as other nearby charges or objects. It is important to carefully control these factors in order to obtain accurate results.

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