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BartSMP [9]
3 years ago
12

Charges of 7.2nC and 6.7nC are 32 cm apart. Find the equilibrium position for a -3.0nC charge.

Physics
1 answer:
dolphi86 [110]3 years ago
3 0

Answer:

I don't know I'm sorry I will tell you another answer asks me

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Climate is determined by averaging the seasonal weather conditions for a region over a period of many ______ years

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A sailfish swims 120 km/hr. How far will it travel in 8.0 minutes?
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16km

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Why is it important to know the direction of the force applied to a moving object and the direction in which the object is movin
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When Mendeleev organized elements in his periodic table in order of increasing mass, similar elements with similar properties we
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8 0
3 years ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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