Expand your knowledge base with the help of IDNLearn.com's extensive answer archive. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.
Sagot :
The charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/C is -2.19C
A particle with some charge moving in an electric field experience force due to electric field.
For the charged particle to remain stationary the force due to electric field and the weight of the particle should be equal.
Force due to electric field = QE
where, Q is the charge of the particle
E is the electric field = 660N/C
Weight of the particle = mg
where, m is the mass of particle = 1.45g = 0.00145kg
g is the acceleration due to gravity = 10m/s²
Hence, force due to electric field = weight of the particle
QE = mg
⇒ Q × 660 = 0.00145 × 10
⇒ Q = 2.19 C
Since the electric field is downward, hence to balance the charged particle, the force applied should be in the upwards direction. So, this is possible if the charge Q is negative.
So, the charged particle has a charge of -2.19C
Learn more about Force here, https://brainly.com/question/28042581
#SPJ4
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.
