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A charge of 2 c is at the origin. when charge q is placed at 2 m along the positive x axis, the electric field at 2 m along the negative x axis becomes zero. what is the value of q?

Sagot :

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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