What is the magnitude of the electric field 27.6 cm directly above an isolated 3.50×10−5 C charge? What is the direction of the electric field?
Electric field of a charge Q at a distance r from it is given by
E = kQ/r2
E = (9*109) * (3.50*10-5C) / (27.6*10-2m)2
E = 4.135*106 N/C
electric field is directed vertically upwards.
What is the magnitude of the electric field 27.6 cm directly above an isolated 3.50×10−5 C...
A) What is the magnitude of the electric field 25.5 cm directly above an isolated 3.50×10−5 C charge? B) What is the direction of the electric field?
What is the magnitude of the electric field 38.1 cm directly above an isolated 3.50×10−5 C charge?
What is the magnitude of the electric field 23.4 cm directly above an isolated 3.50×10−5 C charge?
What is the magnitude of the electric field 21.3 cm directly above an isolated 2.93×10−5 C charge? What is the direction of the electric field?
a) What is the magnitude of the electric field 31.8 cm directly above an isolated 2.71×10−5 C charge? b) What is the direction of the electric field? What is the direction of the electric field? upward downward
What is the magnitude of the electric field 31.8 cm directly above an isolated 2.05×10−5 C charge?
Problem 16.21 What is the magnitude of the electric field 19.2 cm directly above an isolated 2.27×10−5 C charge? Thank you
What magnitude charge creates a 1.60 N/C electric field at a point 3.50 m away?
The electric field in units of N/C at a distance of 26 cm from an isolated point particle with a charge of 2 *10-9 C is:
A particle has a charge of -3.10 nC . A.)Find the magnitude of the electric field due to this particle at a point 0.235 m directly above it. B.) Find the direction of the electric field at this point. (upward or downward) C.) At what distance from the particle does its electric field have a magnitude of 10.5 N/C ?