Physics, 22.08.2019 05:10 laqu33n021

Four charges are on the four corners of a square. q1 = +5μc, q2 = -10μc, q3 = +5μc, q4 = -10μc. the side length of the square is 3 m. what is the magnitude of the net electric field at the center of the square?
show the equation(s) used and your work.

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Answer from: ansley81

Explanation:

Electric field due to a point charge Q at a point at distance d is given by the relation

E = $\frac{K\times Q}{d^2}$

Since Q1 and Q2 are of the same magnitude and distance , so they will create eletric field of same magnitude. Similarly field due to rest of the charges will also be same.

The charges are situated on the corners of a square in such a way that

equal charges of Q1 and Q3 are situated on the diametrically opposite corners of the square. Fields due to these two charges will be equal and opposite in direction. Therefore net field due to these two charges will be zero.

On the same ground, we can say that field due to Q2 and Q4 at the centre will be equal and opposite and therefore they will cancel out each other. Net field at the centre will be zero

Overall, net field due to all the four charges will be zero

Answer from: Quest

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Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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Four charges are on the four corners of a square. q1 = +5μc, q2 = -10μc, q3 = +5μc, q4 = -10μc. the...

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