Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
