00:01
Here we have to solve this problem and first in question a we have to find magnitude in the direction of the electric field at 0 .0 so let's label the charges as 1 and 2 both charges are positively charged that's why charge 1 will create the field 2 the right charge 2 will create the field to the left so the overall charge will be directed to the right, to the left, then.
00:36
And this e equals to negative e2 plus e1, which is negative k, which is negative k times 2k0, divided by d squared minus plus kq0 divided by d squared that equals to negative kq0 squared square it over d squared so that is the answer to question a so it means that this factor is to the left is directed to the left now let's answer question b and here we have to find the potential at this point 0.
01:54
So the potential equals to a superposition of two potentials, which is kq0 over d times 2 plus oh sorry kq0 over d plus k times 2 q0 over d.
02:13
That equals to 3 kq0 over d.
02:19
Now let's answer question 3.
02:24
Here we have to find an external work to bring a negative charge from infinite point to here.
02:33
So delta pe equals to q deltive.
02:44
And here we have to find the external work.
02:54
So on here we bring negative charge to the field created.
03:07
By the positive charges.
03:13
Then we can find absolute value of this work first.
03:19
So first this work equals to absolute value of the potential at this point times the unit charge, which is 3kqq0 over d.
03:48
And now we have to determine have to determine the sign and basically this work should be 3kqq0 over d...