🇳🇬 There should be a mistake in this calculation. I stand to be corrected, though. When resolving vectors. The horizontal component should be Fcosø while the vertical component should be Fsinø. In this case, there is no x component because it cancels out. And the electric field at the center would, therefore, become the y component of the resolved fields. Since E = X + Y, i.e., E = 0 + Y. Which means that a test charge placed at the center of the square would experience an upward force.
hello I have a suggestion that might make it more simple without having to use cos or x and y component. You could apply pythagorean theorum on one of the triangles of this square. In that way you don' t have to devide the length in 2 and wouldn't have to use a 45 degree angle or cos.
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Please make more videos…Interest is growing and your explanations are really good!…Thank you! 🇺🇦🇨🇦
Yep. Brings back memories. E&M at engineering school.. 1980. GO Univ. Alaska, Fairbanks|
im so cooked ;/
Cleared my confusion. Thanks.
🇳🇬 There should be a mistake in this calculation. I stand to be corrected, though. When resolving vectors. The horizontal component should be Fcosø while the vertical component should be Fsinø. In this case, there is no x component because it cancels out. And the electric field at the center would, therefore, become the y component of the resolved fields. Since E = X + Y, i.e., E = 0 + Y. Which means that a test charge placed at the center of the square would experience an upward force.
thank u ! rly clean to look at and very understandable !
You helped me a lot in applied physics respect from Pakistan ❤
Happy to help
hello I have a suggestion that might make it more simple without having to use cos or x and y component. You could apply pythagorean theorum on one of the triangles of this square. In that way you don' t have to devide the length in 2 and wouldn't have to use a 45 degree angle or cos.
Good evening Sir
Could you please upload the video of Electric Field and Electric Field Intensity?
Topic: Dielectric and Relative permittivity
very helpful. thanks
One doubt.... answer please.....
q3 is negative 5 and q2 is negative 5 .,then what is the direction of force on q2 due to q3 ?
what about for a parallelogram?
Plz there in total electric field more explaination
correction for formula k= 9x10^9 N m^2/C^2
I thought we use sin to find y-components ?
Ye
exactly my question
Thanks sir and love from Pakistan 🇵🇰
Just brilliant man
Good work ☺️👍
Also 500cm is 5m not 0.05
thank you
Nice!
those who r preparing for jee 🌚🌝sees this question in exam
Good video from pakistan❤
Bloody legend!