Electric field strength due to two point charges

[stunner5000pt]

Given are two charges +1 X 10^-6 and one 3 X 10 ^ -6 distnace of 10 cm apart

I know there is a poitn in between these two and it is 4 cm from the 1.0 X 10^-6 charge

But i am asked to plot the elctric field strength as a function
if the elctric field is pointing right, then positive, otherwise if poitning left, then negative

Let what happens on the outside of these two charges be, i can figure that on my own

But what happen in the middle


Point at 4 cm and rises after??

(+)------------- (+)

Negative

 

[nrqed]

Given are two charges +1 X 10^-6 and one 3 X 10 ^ -6 distnace of 10 cm apart

I know there is a poitn in between these two and it is 4 cm from the 1.0 X 10^-6 charge

3.66 cm, you mean? (It is not 4.00 cm for sure so I assume you rounded off)

But i am asked to plot the elctric field strength as a function
if the elctric field is pointing right, then positive, otherwise if poitning left, then negative

Let what happens on the outside of these two charges be, i can figure that on my own

But what happen in the middle


Point at 4 cm and rises after??

(+)------------- (+)

Negative

I hope you only have to roughly sketch the result because it's not a simple function to draw at all. But it starts at + infinity near the left charge, decreases sharply, crosses into the negative values as you pass the 3.66 cm mark (assuming the left charge is the 1 microcoulomb), goes down and approaches -infinity near the right charge.

Unfortunately, I have to leave (it's midnight here and I am still in my office. And I am teaching tomorrow at 8h30!).

Good luck with your questions!

Pat

 

[M98Ranger]

(-) | 4 cm | (+)

The electric field strength at the point 4 cm from the +1 X 10^-6 charge and 6 cm from the +3 X 10^-6 charge can be calculated using the formula:

E = k*q/r^2

Where:
E - electric field strength
k - Coulomb's constant (9x10^9 Nm^2/C^2)
q - charge of the point charge
r - distance between the point charge and the point at which the electric field is being measured

In this case, the electric field at the point 4 cm from the +1 X 10^-6 charge and 6 cm from the +3 X 10^-6 charge can be calculated as:

E = (9x10^9 Nm^2/C^2) * (1 X 10^-6 C) / (0.04 m)^2 = 562.5 N/C

Since the two charges are of opposite signs, the electric field at this point will be directed towards the +3 X 10^-6 charge, making it positive. This means that the electric field strength is pointing to the right.

On the other hand, if we consider a point between the two charges, for example at 5 cm from the +1 X 10^-6 charge and 5 cm from the +3 X 10^-6 charge, the electric field strength can be calculated as:

E = (9x10^9 Nm^2/C^2) * (1 X 10^-6 C + 3 X 10^-6 C) / (0.05 m)^2 = 2400 N/C

Since the two charges are of the same sign, the electric field at this point will be directed towards the +1 X 10^-6 charge, making it negative. This means that the electric field strength is pointing to the left.

In general, the electric field strength will decrease as we move away from the charges and will depend on the distance and the magnitude of the charges. At points further away from the charges, the electric field strength will approach zero.