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In the figure above, four charges are arranged. If the magnitudes of all the charges q are all the same and the distance r between them is as shown above, what is the magnitude of the net force on the bottom right charge in terms of $q, r,$ and $k\left(\text { where } k=\frac{1}{4 \pi \varepsilon_{0}}\right) ?$(A) $k\left(\frac{q^{2}}{2 r^{2}}\right)(1+\sqrt{2})$(B) $k\left(\frac{q^{2}}{r^{2}}\right)(1+\sqrt{2})$(C) $k\left(\frac{q^{2}}{2 r^{2}}\right)$(D) $k\left(\frac{q^{2}}{r^{2}}\right)$

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Chapter 12

Practice Test 2

Section 1

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Hi. In the given problem there is a square having side land are each and a charge less skill has been kept at each of its corner. No remain name. This is square to be mm B and see and deep. The length of journal will be equal to side into rules too. This is our route to. Now. We have to find networks experienced by one of the charge kept at the lower right corner of the square. So this Charge will experience three forces at all. One of the force experienced by this is F. F. C. Due to D. The repulsion. Another question is due to the charge put at the so this is F. Etc. Due to be and one more repulsion. Is this due to the charge you put at A. This is exceed you do A. Now this F. F. C. Due to A may be resolved into two components as this angle will be 45 degrees. So one of the components here, this will be F at sea due to a because 45 degree and another component. This FFC due to a shine 45°. Now, first of all, we find the values of FCD and FCB. And actually these two forces will be equal to each other. Using columns law. This escape into cuba into Q two as both are same. So this will be Q squared divided by the square of distance between them. K. Q square by square. Then we find the value of this F. At sea due to a using same columns. Lotuses came into cuba into Q. Two. Again the sylvie Q squared divided by square of distance between them, which is actually the length of diagonal. So it will come out to be a K. Q square by two are square. So overall force along X axis will be F C U D plus horizontal component of this Fc, which is F C. Cause 45°. So it will become okay. Q square by our square plus gay. Do you square by to our square and for the value of course 45° this is one by two. So if we take this K Q square by square is a comin out, leaving behind one, bless one by two. And similarly the same will be the value of networks along Y axis. Hence the net force acting on this charge particle will be the square root off the addition of X squared plus F Y squared. But as we have seen, F Y square is also equal to f X square. So it will be equal to to wise off F X squared. So we can see this is F X times route to. Hence this horse now will be given by for efforts. This is Kay into Q square by square in two, one plus 1 x 2, 2 multiplied by two. So here it will be kick you square by our square into route to Plus one x 2. So if I take one x 2 as a comin out, it will be KQ square by the what is flat and in the private it will be 2 2 Plus one. Hence, we can see here corruption. If we look into the options, none of the option is coming to be correct. So if we consider a misprint option, a maybe taken us correct. Thank you.

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