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Luba_88 [7]
3 years ago
15

Based on these answers how can I determine when electric field strength is affected or not

Physics
1 answer:
Stolb23 [73]3 years ago
6 0

It's not that hard. Let me explain...

There are two ways to remember this.

The first one is by using the formula for the electric strength (E)

E = k × (Q / r^2) , where:

  • k is a constant
  • Q is the "central" charge. The one producing the electric field. <u>Not</u> the one we use to test it.
  • r is the distance between the charges

As you can see, field's strength is proportional to the magnitude of Q, and inverse proportional to the square of the distance. The field does not care how much you alter the charge you positioned in ot to test it.

Although that is the only way to work out the problem, there are many ways to remember it. Given that i have difficulty in remembering formulas, i always try to derive it from something i know.

By definition: E is equal to the force acted on the test charge, divided by its value.

E = F / q

  • F --> Coulomb's force
  • q --> test charge

Since i remember that:

F = k ( Q × q) / r^2

just take out from the expression the q term, since you divided by it, and all it remains is the strength of the field.

In other words, the strength changes as long as something from the expression is altered.

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F_a=1470\ N

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<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

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\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

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\displaystyle F_a=F_{r1}+F_{r2}.....[1]

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\displaystyle F_{r2}-T=0

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\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

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