Maximizing Electrostatic Force between Two Point Charges

In summary: I already explained why I think we can leave out the constants.Differentiation wrt a is allowed because we are trying to find a value for ##\lambda## that gives the maximum ##F##.
  • #1
Vishakha
12
0

Homework Statement


Two point charges q and λq located at the points, x=a & x=μa respectively. If the sum of the two charges is constant,what is the value of λ for which the magnitude of the electrostatic force is maximum?

Homework Equations


The Attempt at a Solution


For force to be maximum dF/dq =0 and d^F/dq^2 <0. When I tried to calculate dF/dq =0 I got λ=0.
 
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  • #2
Well, how did you do that? Can you post your steps in detail ?

By the way: why calculate ##dF\over dq## ?
 
  • #3
BvU said:
Well, how did you do that? Can you post your steps in detail ?

By the way: why calculate ##dF\over dq## ?

I already wrote that in The attempt at a solution section.

F= λq2/4πε0 (μa-a)2
⇒ dF/dq = 2λq/4πε0(μa-a)2 =0
⇒ 2λq = 0
q≠0 ⇒λ=0
 
  • #4
If you want to find the maximum of a function of x you take the derivative wrt x.
Here, do you want to consider the force as a function of q ?

Note also that you forgot to make use of the given that
Vishakha said:
the sum of the two charges is constant
 
  • #5
BvU said:
If you want to find the maximum of a function of x you take the derivative wrt x.
Here, do you want to consider the force as a function of q ?

Note also that you forgot to make use of the given that
If I used q+λq = C ⇒ λq=C-q
Then 2λq=0 ⇒ 2(C-q) = 0 ⇒q=C

Let F is function of distance

F = λq2/4πε0a2(μ-1)2
⇒ -2λq2/4πε0a3(μ-1)2 = 0
⇒ -2λq2 = 0
⇒ λq= 0 or λq=C
 
  • #6
Are you saying ##\lambda = 0 \Rightarrow F = 0 ## too ? That is easily proven wrong !
 
  • #7
BvU said:
Are you saying ##\lambda = 0 \Rightarrow F = 0 ## too ? That is easily proven wrong !

You are right. F shouldn't be zero but I don't find any mistake in my calculation.
 
  • #8
Vishakha said:
I don't find any mistake in my calculation
I did and I tried to point it out. You want to express F in terms of ##\lambda## and differentiate wrt ##\lambda##. Make a start ...
 
  • #9
BvU said:
I did and I tried to point it out. You want to express F in terms of ##\lambda## and differentiate wrt ##\lambda##. Make a start ...

You mean I have to differentiate F wrt λ and distance between charges and q is constant.
 
  • #10
Yes and No. In that order:
Vishakha said:
the sum of the two charges is constant
 
  • #11
BvU said:
Yes and No. In that order:

λ+qλ = C
⇒dq = -(q+1)dλ/λ ... (1)

dF/da = [(μa-a)2 { 2λq dq + q2 dλ} - 2q2λa (μ-1)2 ]/ 4πε0 (μa-a)2 = 0

After putting value of eq (1) I got final eq
-q(q+1) dλ = 2λa
 
Last edited:
  • #12
Can you explain why you are now differentiating wrt a ? I thought we agreed to seek a value for ##\lambda## that gives the maximum ##F## ?

And: do you think we can leave out the constants ##\displaystyle {1\over 4\pi\varepsilon_0 (\mu a - a)^2 } ## ?
 
  • #13
BvU said:
And: do you think we can leave out the constants ##\displaystyle {1\over 4\pi\varepsilon_0 (\mu a - a)^2 } ## ?

I got λ=1 if we leave the constants.

But I don't understand why are we differentiating wrt λ?
 
  • #14
Vishakha said:
I got λ=1 if we leave the constants.

But I don't understand why are we differentiating wrt λ?
Because the problem asks, "what is the value of λ for which the magnitude of the electrostatic force is maximum?"
 
  • #15
Vishakha said:
I got λ=1 if we leave the constants.

But I don't understand why are we differentiating wrt λ?
Can you explain how you found ##\lambda = 1 ## ?
And can you justify leaving out the constants ? Why is that allowed ?
 

1. What is electrostatic force?

Electrostatic force is the force between electrically charged objects that results from the attraction or repulsion between their charges.

2. What are the factors that affect electrostatic force?

The factors that affect electrostatic force include the magnitude of the charges, the distance between the charges, and the medium between the charges.

3. How can we calculate electrostatic force?

The electrostatic force between two charged objects can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

4. What is the SI unit of electrostatic force?

The SI unit of electrostatic force is Newton (N).

5. How does electrostatic force differ from gravitational force?

Electrostatic force is caused by the attraction or repulsion between electric charges, while gravitational force is caused by the mass of objects. Additionally, electrostatic force can be either attractive or repulsive, while gravitational force is always attractive.

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