Chapter 4: Q42E (page 713)

A point charge is placed at each corner of a square with side length a. All charges have magnitude q. Two of the charges are positive and two are negative (Fig. E21.42). What is the direction of the net electric field at the canter of the square due to the four charges, and what is its magnitude in terms of q and a?

Short Answer

Expert verified

Answer:

If a point charge is kept at a corner of a square with side length a, and all charges have the magnitude of q. And if two of them are positive and two of them are negative, therefore the direction of the net electric field at the centre due to the charges due to the four charges is in the direction. And the magnitude in terms of q and a is $\frac{1}{4 π ε_{0}} \frac{8}{\sqrt{2}} \frac{q}{a^{2}}$

Step by step solution

Direction of the charge

It is given that the charges have the same magnitude and the point is at the canter. Hence the charges have the sameand at a same distance of r at the canter. We will calculate E for one charge and then multiply it by.

Now, as the electric fields are in thedirection and cancels each other due to the opposite charges. Hence the electric field will be only in the y direction. Now we getby:

$E_{y} = E c o s 45 = \frac{1}{4 π ε_{0}} \frac{| q |}{r^{2}} c o s 45$

Calculating the magnitude

From the above equation when the term $\frac{1}{4 π ε_{0}}$ equals to $9.0 \times 10^{9} N \cdot m^{2} / C^{2}$

Then we get:

$r^{2} = {(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2} r = \sqrt{{(\frac{a}{b})}^{2}} + \sqrt{{(\frac{a}{b})}^{2}} r = \frac{a}{\sqrt{(2)}}$

Now:

$E_{y} = \frac{1}{4 π ε_{0}} \frac{q}{r^{2}} c o s 45 E_{y} = \frac{1}{4 π ε_{0}} \frac{| q |}{{(\frac{a}{\sqrt{2}})}^{2}} (\frac{1}{\sqrt{2}}) E_{y} = \frac{1}{4 π ε_{0}} \frac{2}{\sqrt{2}} \frac{q}{a^{2}}$