Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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 appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
