Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
twitter
Group Title
Electric field strength is defined: “force per unit positive charge on a small test charge”.
Why is it necessary for the test charge to be small?
 one year ago
 one year ago
twitter Group Title
Electric field strength is defined: “force per unit positive charge on a small test charge”. Why is it necessary for the test charge to be small?
 one year ago
 one year ago

This Question is Open

Krishnadas Group TitleBest ResponseYou've already chosen the best response.0
Because if the test charge was too large,the effect by the test charge on source charge will be high.So there is a chance that the source charge will tend to move.
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
who said the test charge is small???? +1C charge is very big charge missy!.. we consider a UNIT charge.. something that has a magnitude one in their respective units.. in SI that unit turns out to be a COULOMB.. and what krishnadas said is rubbish :P.. we need to define field at a point in some way.. so we need a reference.. its easy to remember 1C than remembering a mili coulomb or a micro coulomb.. !!..
 one year ago

Carl_Pham Group TitleBest ResponseYou've already chosen the best response.0
The test charge generates its own electric field, which adds (vectorially) to the field you're trying to measure. The force on the test charge is the result of the sum of the fields. The only way you can disentangle the result of the field you're trying to measure is to make measurements as you reduce the size of the test charge, and extrapolate the result to charge > 0. That's the process described, in a shorthand way, by saying you use a small test charge. It means you take the q>0 limit.
 one year ago

gleem Group TitleBest ResponseYou've already chosen the best response.0
Mashy is correct, the size of the test charge is irrelevant. The idea is to determine the electric field by determining the force that it exerts on the test charge as given by Coulombs's Law. Beside the field of the test charge no matter how small is infinite at that point and would essentially wipe out the field from the other charge making the measurement impossible. the electric fields do not interact. The charges interact with the others field.
 one year ago

gleem Group TitleBest ResponseYou've already chosen the best response.0
Question. From a practical point of view if the source charge is distributed and mobile say on a large conducting sphere and the test charge where large enough or close enough could cause a redistribution of the charge on the sphere thus altering the original electric field ?
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
Test charge is JUST A CONCEPT.. you don't really have to PUT a charge their.. its like WHAT IF there was a charge there.. and WHAT IF it had no effect.. on the SOURCE.. well obviously in reality the source is affected. but when i wanna describe field due to just one charge.. (source).. thats the only way to describe it..
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.