Asap help please~electric fields from continuous charge distribution?

In summary, the conversation discusses finding the expression for the x-component of the electric field at a point above one end of a nonconducting rod with uniformly distributed charge. The answer involves integrating a function and setting limits for the variables, with the final expression being [kQ/(2a^2*sq rt(5))]*{sq rt(5) -1}. The conversation ends with the confirmation that the page provided was helpful in understanding the concept.
  • #1
Major Mei
4
0
Hello guys,

I tried to figure this out and I got my answer. I just want to check it. So would you guys please help me with it? Thank you!
Here is the question:
A nonconducting rod of length 2a has a charge Q uniformly distributed along it. Find the expression for x-component of the electric field at point P(a distance above one of the ends of the rod)

my answer:

Ex = Integral from x = -2a to 0 of [(k*Q/2a)*xdx/ (x^2+a^2)^(3/s2)]
set x^2+a^2 = y
x*dx = dy/2 also x = -2a y = 5a^2
x = 0 y =a^2

Ex = -2*(k*Q/4a)[1/sq rt(y)] within limits of y = 5a^2 to y = a^2]
= - (kQ/2a)*[1/sq rt(5a^2) – 1/sq rt(a^2)] = -(kQ/2a^2)[1/sq rt(5) -1]
= [kQ/(2a^2*sq rt(5))]*{sq rt(5) -1]

is that correct?
Thank you and I am willing to learn from you ;)
 
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  • #3

Related to Asap help please~electric fields from continuous charge distribution?

1. What is an electric field from a continuous charge distribution?

An electric field from a continuous charge distribution is a region in space where a collection of charges exert a force on other charges placed within that region. It is a vector field, meaning it has both magnitude and direction.

2. How do you calculate the electric field from a continuous charge distribution?

To calculate the electric field from a continuous charge distribution, you can use Coulomb's law and the principle of superposition. This involves breaking down the charge distribution into infinitesimal elements and calculating the electric field due to each element, then adding them together to find the total electric field.

3. What is the unit of electric field from a continuous charge distribution?

The unit of electric field from a continuous charge distribution is newtons per coulomb (N/C), or volts per meter (V/m). This unit represents the force per unit charge that a charge experiences when placed in the electric field.

4. How does the electric field from a continuous charge distribution affect charged particles?

The electric field from a continuous charge distribution can exert a force on charged particles placed within the field. The direction of the force depends on the sign of the charge and the direction of the electric field. Positive charges will experience a force in the same direction as the electric field, while negative charges will experience a force in the opposite direction.

5. Can the electric field from a continuous charge distribution be negative?

Yes, the electric field from a continuous charge distribution can be negative. This means that the force on a positive charge would be in the opposite direction as the electric field, and the force on a negative charge would be in the same direction as the electric field. Negative electric fields can occur when there is a distribution of negative charges or a distribution of positive charges with a stronger magnitude.

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