Electrostatics help please -- Electric field, potential

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Two spheres, one with a charge of -20 mC and the other with -50 mC, are positioned 5 meters apart, and the discussion focuses on calculating the electric field and potential at the halfway point. The relevant equations for the electric field from each sphere are provided, with suggestions to clarify the use of negative charges and to utilize LaTeX for better readability. Participants emphasize the importance of calculating the electric fields separately and considering their directions, as they may oppose or reinforce each other. Additionally, the formula for electric potential is mentioned, prompting further calculations for the potential at the halfway point. The conversation highlights the need for clear communication and understanding of the underlying physics principles.
RamirezA
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Homework Statement
Calculate the electric field (halfway point) and value of the electric potential at the halfway point
Relevant Equations
E=kQ1/r1^2
E=kQ2/r2^2
E=E1+E2
Hello, any answers appreciated:
'Two spheres are 5 m apart. Sphere 1 has a charge of -20 mC and sphere two has a charge of -50 mC. (a) Find the strength of the electric field at the sphere's halfway point. (b) Find the electric potential at the halfway point
 
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RamirezA said:
Homework Statement:: Calculate the electric field (halfway point) and value of the electric potential at the halfway point
Relevant Equations:: E=kQ1/r1^2
E=kQ2/r2^2
E=E1+E2

Hello, any answers appreciated:
'Two spheres are 5 m apart. Sphere 1 has a charge of -20 mC and sphere two has a charge of -50 mC. (a) Find the strength of the electric field at the sphere's halfway point. (b) Find the electric potential at the halfway point
Urgent or not, forum rules require you to post an attempt.
 
haruspex said:
Urgent or not, forum rules require you to post an attempt.
E1= (9x10^9)(10x10^-6)/(5)^2 (dont know where/if too include the negative values in this)
= 3600
E2= (9x10^9)(30x10^-6)/(5)^2
=10800
=14400
 
RamirezA said:
E1= (9x10^9)(10x10^-6)/(5)^2 (dont know where/if too include the negative values in this)
= 3600
E2= (9x10^9)(30x10^-6)/(5)^2
=10800
=14400
I don't claim to be any expert but I will try to give you a few hints.

First try using latex for your formulas/equations it is quite difficult to read without it,and a lot of people won't even bother reading through your question if you write it out like that.

Second the formula for the electric field is this;

$$ E = \frac{Q}{4\pi \epsilon0 r^2} $$ I am guessing the k in your formula is just the pi and epsilon put in one constant;

Now I am going to assume that you know what all of these are; that Q is your charge pi the value of pi (3.14) etc..

Now the distance between your spheers is 5 m, if you are asked to calculate the strength of the E field at the halfwaypoint,what is your distance then?

Also the - infront of your spheers mean that they are negative charged.

What I would do if I were you (for part a) is calculate E1 and E2 at the halfwaypoint and than try to get the ##E_{total} ##.

For all the homework helpers,if I am leading Ramirez astray,let him (and me) know,I am just going of the litle knowledge I have.
 
arhzz said:
I don't claim to be any expert but I will try to give you a few hints.

First try using latex for your formulas/equations it is quite difficult to read without it,and a lot of people won't even bother reading through your question if you write it out like that.

Second the formula for the electric field is this;

$$ E = \frac{Q}{4\pi \epsilon0 r^2} $$ I am guessing the k in your formula is just the pi and epsilon put in one constant;

Now I am going to assume that you know what all of these are; that Q is your charge pi the value of pi (3.14) etc..

Now the distance between your spheers is 5 m, if you are asked to calculate the strength of the E field at the halfwaypoint,what is your distance then?

Also the - infront of your spheers mean that they are negative charged.

What I would do if I were you (for part a) is calculate E1 and E2 at the halfwaypoint and than try to get the ##E_{total} ##.

For all the homework helpers,if I am leading Ramirez astray,let him (and me) know,I am just going of the litle knowledge I have.
Thanks man, that helps
 
RamirezA said:
Thanks man, that helps
Consider in what direction each field acts. Do they oppose each other or reinforce?
For part b, what is the formula for potential?
What answers do you get?
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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