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N76 [4]
3 years ago
6

Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values

provided below and record the resulting forces.
Left Charge Righ Charge Resulting force(N)
1μC 4μC
4μC 1μC
2μC 2μC
1μC 2μC
1μC 8μC
2μC 8μC
Physics
1 answer:
Vlad1618 [11]3 years ago
5 0

Answer:

Explanation:

Force between two charges can be expressed as follows

F = k q₁ q₂ / d²

q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹

distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m

force between 1μC and  4μC

= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 4μC and  1μC

= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 2μC and  2μC

= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 9 x 10 = 90 N

force between 1μC and  2μC

= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²

= 4.5 x 10 = 45 N

force between 1μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 18 x 10 = 180 N

force between 2μC and  8μC

= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²

= 36 x 10 = 360 N

Left Charge   Right Charge Resulting force(N)

1μC                     4μC                  90 N

4μC                   1μC                    90 N

2μC                  2μC                    90 N

1μC                    2μC                   45 N

1μC                  8μC                    180 N

2μC                  8μC                  360 N

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Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
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Answer:

(a) \vec{R}= 4.83\ m\ \hat{i}+1.47\ m\ \hat{j}

(b) (5.05 m, 16.93 degrees wrt x-axis)

Explanation:

Given:

  • \vec{D} = (3.00 m, 315 degrees wrt x-axis)
  • \vec{E} = (4.50 m, 53.0 degrees wrt x-axis)

Let us first fond out vector D and E in their rectangular form.

\vec{D} = (3\cos 315^\circ\ \hat{i}+3\sin 315^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.12\ \hat{i}-2.12\ \hat{j})\ m\\

Similarly,

\vec{E} = (4.5\cos 53^\circ\ \hat{i}+4.5\sin 53^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.71\ \hat{i}+3.59\ \hat{j})\ m\\\because \vec{R}=\vec{D}+\vec{E}\\\therefore \vec{R} = (2.12\ \hat{i}-2.12\ \hat{j})\ m+(2.71\ \hat{i}+3.59\ \hat{j})\ m\\\Rightarrow \vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m

Part (a):

We can write the resultant vector R as below:

\vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m

Part (b):

Magnitude\ of\ resultant = \sqrt{4.83^2+1.47^2}\ m = 5.05\ m\\\textrm{Direction in angle with the x-axis} = \theta = \tan^{-1}(\dfrac{1.47}{4.83})= 16.93^\circ

Since both the components of the resultant lie on the positive x and y axes. So, the resultant makes an acute angle with the positive x-axis.

So, R = (5.05 m, 16.93 degrees wrt x-axis)

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