Fleid 01 forces like the electrical force, we often talk about a field that is s

Question

Fleid 01 forces like the electrical force, we often talk about a field that is set up around a charged object or a group of charged objects. Once we know the strength of the When studying electric field at a location in space, we can figure out the force another charged partidle would experience if put at that point in space. To calculate the force, you simply mulbiple the electrical charge of the object by the strength of the electric field (F Since a field is a vector (it has both a magnitude QE) and a direction), we also have to decide how to determine the direction of the force on a small positively charged particle placed at that point in space. If the field points to the North, a tiny positively charged particle (say a proton) would experience a force to the North. A negatively charged particle, for example an electron, would experience a force to the the direction of the field. By convention the direction of the Beld at a point in space is South, in the direction opposite the direction of the fleid. electric field of 4.06x 103 N/C which points to the right. If you put an object weh acharge of +0.256 C at that point in space, Suppose at some point in space there is an and direction of the electrical force on that object? D to the leftto the right Submit Answer Tries 0/10 point in space, what is the magnitude and direction of the electrical force on that object to the left to the right bit nswer Tries 0/10 experimet you place an object with an electrical charge of +0.360 C at a point in space, direction of the to the left to the right MacBook Air 1 LA/

Explanation / Answer

Solution :

We have electric field E at some point is 4.06 * 103 N/C towards right

a) charge, Q = 0.256 C

Therefore force acting on charge Q = 4.06 * 103 *0.256 N = 1.03936 * 103 N towards right bacause there is sign change in force with respect to electric field

b) charge Q = -0.256 C

Force on charge Q = 4.06 * 103 * 0.256 = 1.03936 * 103 N towards left sign of force is negative with respect to electric field.

C) we have

Charge Q = 0.360 C

Force experienced F = 725 N

Therefore electric field E = 725/0.360 = 2013.889 N/C towards left because electric field and force have same sign.

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