Two point charges of magnitude +7.00 μC and -9.00 μC are placed along the x-axis at x = 0 cm and x = 40.0 cm, respectively. Where must a third charge, q, be placed along the x-axis so that it does not experience any net force because of the other two charges?

Respuesta :

Answer:

third charge must be placed at x = -298.11 cm

Explanation:

When two opposite charges are placed in a straight line then the electric field is always zero at a point near to the charge of lesser magnitude

so here let say the point where electric field is zero lie on the left side of 7.00 micro C charge at a distance "x" from it

so we will have net electric field due to both charges will be zero at that point

so electric field of charge 1 must be equal to the electric field of charge 2

[tex]\frac{kq_1}{x^2} = \frac{kq_2}{(L + x)^2}[/tex]

so we have

[tex]\frac{7\mu C}{x^2} = \frac{9 \mu C}{(40 + x)^2}[/tex]

now square root both sides

[tex]\frac{2.645}{x} = \frac{3}{40 + x}[/tex]

[tex]105.83 + 2.645 x = 3x[/tex]

[tex]105.83 = 0.355 x[/tex]

[tex]x = 298.11 cm[/tex]

so third charge must be placed at x = -298.11 cm

The third charge, q, should be placed along the x-axis to not experience any net force because the two point charges at 298.11 cm.

What is electric charge?

When a body is places inside a electric field, it fill a force due to the electric field. This force felt by body is called the electric charge. It can be either positive or negative.

The charge of point one is of magnitude +7.00 μC and the charge of point two is of magnitude -9.00 μC.

The point one is placed along the x-axis as, x = 0 cm and point 2 is placed along the x-axis as, x = 40.0 cm. Thus the value of point one is x cm and the value of point two is 0.4 cm.

From the following equation, the charge and the position can be represented as,

[tex]\dfrac{7}{x^2}=\dfrac{9}{(40+x)^2}\\\left(\dfrac{2.645^2}{x^2}\right)=\left(\dfrac{3^2}{(40+x)^2}\right)\\\left(\dfrac{2.645}{x}\right)^2=\left(\dfrac{3}{(40+x)}\right)^2\\[/tex]

Cancel out the square of both sides as,

[tex]\left(\dfrac{2.645}{x}\right)=\left(\dfrac{3}{(0.4+x)}\right)\\2.645\times40+2.645x=3x\\0.355x=105.83\\x=\dfrac{105.83}{0.355}\\x=298.11\rm cm[/tex]

Hence, the third charge, q, should be placed along the x-axis to not experience any net force because the two point charges at 298.11 cm.

Learn more about the electric charge here;

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