Question

two charges-16 and +4c,are fixed in place and separated by 3.0m. at what spot along a line through the chrages is the net electric field zero? what would be the force on a chargeof +14c placed at this spot?

Answer 1

Answer on Question #67475- Physics / Other

Two charges $q_{1} = -16 \, \text{C}$ and $q_{2} = +4 \, \text{C}$, are fixed in place and separated by $d = 3.0 \, \text{m}$ at what spot along a line through the charges is the net electric field zero? what would be the force on a charge of $+14 \, \text{c}$ placed at this spot?

Solution:

Let us assume that zero net electric field is occurring at the point A. So

$$E _ {1} = E _ {2}.$$ $$E _ {1} = k \frac {\left| q _ {1} \right|}{(d + x) ^ {2}}, \quad E _ {2} = k \frac {\left| q _ {2} \right|}{x ^ {2}},$$

We obtain

$$\frac {\left| q _ {1} \right|}{(d + x) ^ {2}} = \frac {\left| q _ {2} \right|}{x ^ {2}},$$ $$x \sqrt {\left| q _ {1} \right|} = (d + x) \sqrt {\left| q _ {2} \right|},$$ $$x \left(\sqrt {\left| q _ {1} \right|} + \sqrt {\left| q _ {2} \right|}\right) = d \sqrt {\left| q _ {2} \right|},$$ $$x = d \frac {\sqrt {| q _ {2} |}}{\sqrt {| q _ {1} |} + \sqrt {| q _ {2} |}} = d \frac {1}{\sqrt {\frac {| q _ {1} |}{| q _ {2} |}} + 1} = \frac {3}{\sqrt {4} + 1} = 1 \, \text{m}.$$

Thus, at the point A there is a zero net electric field.

The force that would be acted on the charge placed at this point is equal to zero.

**Answer**: The zero net electric field there is at the point that displaced from positive charge to the right on $1 \, \text{m}$.

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