Calculating Electric Field due to a Dipole

In summary, the problem involves two dipoles with charges +q and -q, separated by a distance s and located a distance d from point A. Using the equations for the electric field of a dipole, the x component was determined to be 166 N/C and the y component was determined to be 333 N/C. However, due to the arrangement of the dipoles, the net electric field at point A is the sum of these components, resulting in a direction towards the dipoles.
  • #1
guitarman
23
0

Homework Statement


Two dipoles are oriented as shown in the diagram below. Each dipole consists of two charges +q and -q, held apart by a rod of length s, and the center of each dipole is a distance d from location A. If q = 4 nC, s = 1 mm, and d = 6 cm, what is the electric field at location A?

As I am unaware of how to attach a file, here is a diagram
************+
************-
************|
************|
************|
************|
-***********|
+--------------A

the *s are there for formatting, or the else lines do not align

Homework Equations


(1/4*pi*epsilon)*(2qs/d^3) is the electric field of a dipole, on the dipole axis
and
(1/4*pi*epsilon)*(qs/d^3) is the electric field of a dipole, perpendicular to the dipole axis

The Attempt at a Solution


Plugging the values given, I used the latter equation for the x component, and by doing (9e9)(4e-9*1e-3)/(6e-2)^3 I obtained 166 N/C.

I used the first equation for the y component and obtained 333 N/C

I know there is no z component, so I input <166, 333, 0> N/C, yet this is wrong. Can somebody explain to me what I am doing? I attempted submitting -166 for the x component in case I had somehow misinterpreted the direction of the field, but that was also wrong. Does anybody understand what I am doing wrong?
 
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  • #2
In the dipole above A, the negative charge is closer to A . So the attractive force is more than the repulsive force. So the electric field is towards the dipole.
In the equatorial position the direction of the field is parallel to the dipole from positive charge to negative charge.
Hence the electric field due to two dipoles are in the same direction. So the net field is the sum of them.
 
  • #3


Hello,

Thank you for sharing your work and question. Your approach to solving this problem is correct, however, there is a mistake in the values you have used for the length of the dipole (s) and the distance from the dipole to location A (d).

In the problem statement, it is given that s = 1 mm and d = 6 cm. However, in your calculation, you have used s = 1 m (1 millimeter) and d = 6 m (6 meters). This is why your calculated values for the electric field components are incorrect.

To correct this, you can either convert the given values to meters before plugging them into the equations, or you can use the values given in the problem statement and convert the final answer to N/C.

I hope this helps clarify your confusion. Keep up the good work!
 

1. How is the electric field due to a dipole calculated?

The electric field due to a dipole can be calculated using the formula E = (1/4πε0) (q/d2), where E is the electric field, q is the magnitude of the charge, and d is the distance between the two charges. This formula takes into account the Coulomb's law, which states that the electric field is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them.

2. What is the direction of the electric field due to a dipole?

The electric field due to a dipole is a vector quantity, meaning it has both magnitude and direction. The direction of the electric field can be determined using the right-hand rule, where the fingers of the right hand curl in the direction of the positive charge and the thumb points towards the direction of the electric field.

3. How does the distance between the charges affect the electric field due to a dipole?

The distance between the charges, d, has an inverse square relationship with the electric field due to a dipole. This means that as the distance between the charges increases, the electric field decreases. This relationship is explained by the fact that the electric field spreads out over a larger area as the distance between the charges increases, resulting in a weaker electric field.

4. Can the electric field due to a dipole be zero?

Yes, the electric field due to a dipole can be zero under certain conditions. This occurs when the two charges in the dipole are equal in magnitude and opposite in sign, and when the distance between them is infinite. In this case, the electric field vectors cancel each other out, resulting in a net electric field of zero.

5. How does the orientation of the dipole affect the electric field?

The orientation of the dipole can affect the electric field in two ways. First, the direction of the electric field will change depending on the orientation of the dipole. Second, the magnitude of the electric field will be stronger when the dipole is aligned with the electric field lines and weaker when it is perpendicular to the electric field lines.

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