Electrostatics question

kman12

Homework Statement

Hello, 4 charges of equal magnitude are placed on the corners of a square of length a. The two charges at the top of the square are negative, while the two charges at the bottom of the square are positive.

what is the magnitude of electric field at the centre of the square. Ok ive done this part i have [q(2)^(0.5)]/(8.pi.eplion.(0.5)a^2) in a verical (upwards direction).

what direction would an electron accelerate if placed in the centre of the square? im not too sure on this, does an electron acelerate in a direction opposite to the resultant elecric field, since electric field goes from positive to negative.

thanks

Homework Helper
Welcome to PF!

Hi kman12! Welcome to PF!
what direction would an electron accelerate if placed in the centre of the square? im not too sure on this, does an electron acelerate in a direction opposite to the resultant elecric field, since electric field goes from positive to negative.

Yes … electric notation was fixed before electrons were discovered, so electrons always go the "wrong way"!

kman12
cheers,
another quick question, if i want to find the potential across the plates of a capacitor, does it matter in which direction i integrate v= -∫E.ds=-∫ Edscosθ. In other words can i start from the positive end of the capactor and integrate to the negative end. so E and ds are parralal so θ=0 and then flip limits to get positive value for V. OR could i integrate from negative end towards positive end and θ=180 then i dont need to flip limits, i will already have positive v.

Obviously the potential across a capacitor is v=Ed, but im not sure how to arrive at this

Homework Helper
if i want to find the potential across the plates of a capacitor, does it matter in which direction i integrate v= -∫E.ds=-∫ Edscosθ. In other words can i start from the positive end of the capactor and integrate to the negative end. so E and ds are parralal so θ=0 and then flip limits to get positive value for V. OR could i integrate from negative end towards positive end and θ=180 then i dont need to flip limits, i will already have positive v.

Obviously the potential across a capacitor is v=Ed, but im not sure how to arrive at this

Hi kman12!

just got up :zzz: …

Not really following you

if the charge is Q on one side and -Q on the other, then you get there by forcing either Q positive charge one way, or Q negative charge the other way.

For details of the calculation, see the PF Library on https://www.physicsforums.com/library.php?do=view_item&itemid=112"

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Homework Helper
cheers,
another quick question, if i want to find the potential across the plates of a capacitor, does it matter in which direction i integrate v= -∫E.ds=-∫ Edscosθ. In other words can i start from the positive end of the capactor and integrate to the negative end. so E and ds are parralal so θ=0 and then flip limits to get positive value for V. OR could i integrate from negative end towards positive end and θ=180 then i dont need to flip limits, i will already have positive v.

Obviously the potential across a capacitor is v=Ed, but im not sure how to arrive at this
If you're just trying to get the magnitude of the voltage difference (which is usually the case), no it doesn't matter which way you integrate. Magnitudes are always positive, so if your integration gives you a negative answer you just drop the sign (take the absolute value).

When I took introductory electromagnetism as a freshman in college, the professor would use this shortcut all the time - basically the idea was that if you know which sign your answer is supposed to have, there's no need to spend a lot of time figuring out the right way to set up the calculation (e.g. the right direction to integrate). You just pick one way, do the math, and if you get the wrong sign, you flip it. We named it the Verlinde theorem in his honor ("1 = -1 when necessary" )

kman12
o ok thanks guys. I think my lecturer does the same thing by dropping the sign. But sometimes he doesnt even state that the e field and length element ds, are in oppisite directions, ie 180 degrees. It makes sense now cheers.