Calculating Electric Field Between Charged Particles

AI Thread Summary
To calculate the electric field at the midpoint between a -8.0 microcoulomb and a 7.0 microcoulomb charge, the electric fields due to each charge must be determined separately and then combined. The force between the charges is calculated using Coulomb's law, resulting in a force of approximately 78.75 N. The electric field is then found using E=F/q, where the test charge is considered to cancel out in practical scenarios. It's important to account for the direction of the electric fields, as they will affect the net electric field at the midpoint. The correct approach involves calculating the contributions from each charge and adding them vectorially.
jena
Messages
74
Reaction score
0
Hi,

My Question:

What are the magnitude and direction of the electric field at a midway between a -8.0 microcolumb and a 7.0 microcolumb charge 8.0 cm apart? Assume no other charges are nearby.

Work

E=F/q

F=(kQ1Q2)/r^2

F=((9.0 X 10^9 N m^2/C)(-8.0 X10^-16 C)(7.0 X10^-6C))/(.08m)^2

F=-78.75 or 78.75 N

E=(78.75N)/(1.6X10^-19C)

E=4.92 x 10^20 N/C

Is this correct or should I look at findin the electric field using Q1 and Q2 seperately then adding them together to find a net elecrical field.

Thank you :smile:
 
Last edited:
Physics news on Phys.org
You need to find the field at the midpoint due to each charge, then add those fields. Don't forget: Direction matters.
 
Last edited:
In E=F/q , the q has to be the same *test charge* that the Force acts on!
That is, in all practical situations, the test charge q cancels.
The part of the E-field at test location t, contributed by charge Q1, is
E_1t = k Q1/(r_1t)^2 (away),
where r_1t = distance from Q1 to the test location (.04m in your case).
 
Oh... okay thanks :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Back
Top