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# 物理題目電力與電場

Four equal point charges of +3.0µC are placed at the four corners of a square that is 40 cm on a side. Find the force on any one of the charges.

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Apply Coulomb's Law with Superposition.

Coulomb's Law :

F = k * (q1*q2) / r^2 where k = 9x10^9 Nm^2/C^2

Since any point charges are identical to any charge at the square corner, and they are positive charges. Therefore, we can conclude that they will repeal to each other. We can also conclude that the force acts upon any charge is identical.

Let's give each charge a name relative to the square:

q1 = at top right

q2 = at bottom right

q3 = at top left

q4 = at bottom left

For q1, this point charge will repeal due to q2 and q3. So, to find the force on q1, it is the sum of these two forces.

To calculate the force at one corner, first calculate the force acts upon q1 by q2:

F1= (9x10^9) * [ (3.0µ)(3.0µ) / 0.4^2 ] = 0.51 N

Then, calculate the force acts upon q1 by q3:

F2= (9x10^9) * [ (3.0µ)(3.0µ) / 0.4^2 ] = 0.51 N

F1 and F2 are two vectors; therefore, to add them, we need to use vector addition:

Ft = F1 F2 = √{ 0.51^2 0.51^2 } = 0.72 N

Therefore, 0.72 N is the force on any one of the charges. But, since they are vectors, the direction makes all the difference. To give the full answer, we should also include the direction of the force.

Finally,

Ft on q1 = 0.72 N, NW

Ft on q2 = 0.72 N, SW

Ft on q3 = 0.72 N, NE

Ft on q4 = 0.72 N. SE

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This is not an English question. Next time, you should post this type of questions under Physics to increase the change of getting helped.

2008-04-24 13:17:21 補充：

剛剛看了一下數字

有些符號不見了

point charges are 3.0 x 10^-6 不是 3.0 (mu 不見了)

另外 Ft = F1 "plus" F2 的 plus 也不見了

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