Respuesta :
The value of Q will be -8 C.
In the presence of an electric or magnetic field, matter experiences a force due to its electric charge.
A moving electric charge generates a magnetic field, and an electric charge has an accompanying electric field.
The information provided in the issue is;
The separation between and is 2m.
The separation between and is 2m.
An origin charge equals +2 C
The electric fields are identical in magnitude but are facing in different directions. As a result, the following relationship can be used
Q/16=1/2
The value of Q will be -8 C.
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Th value of q is= -8C
How can we calculate the value of the q?
Here, To calculate the value of q we use the formula,
The E-field due to a point charge = [tex]\frac{k\times Q}{D^{2} }[/tex]
Here we are given,
D= the distance along the axes.
k= The coulomb's constant.
We have to find, the charge along the axes = Q
Now, we have to substitute the known values in that equation for the cases,
For the E-field at x=-2 to equal 0, the E-fields from both charges must be equal in magnitude and opposite in direction.
For the first case, E from +2C charge = [tex]\frac{k\times Q}{D^{2} }[/tex] = [tex]\frac{k\times (+2C)}{2^{2} }[/tex]=[tex]\frac{k\times (+2C)}{4 }[/tex].
For the second case, E from Q charge= [tex]\frac{-k\times Q}{D^{2} }[/tex]=[tex]\frac{-k\times Q}{(2+2)^{2} }[/tex]=[tex]\frac{-k\times Q}{16 }[/tex].
Now, Comparing the two cases we can find that,
[tex]\frac{k\times (+2C)}{4 }=\frac{-k\times (Q)}{16}[/tex]
Or,[tex]\frac{+2C}{4}= \frac{-Q}{16}[/tex] (Eliminate k from both sides of the equation.)
Or,[tex]Q=-8C[/tex]
So, from the above calculation we can conclude that,
The value of q is -8C.
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