Can you please solve the blank one Two metal plates separated by 0.45 cm are at

Question

Can you please solve the blank one

Two metal plates separated by 0.45 cm are at potentials of -225 V and +550 V respectively. Draw a diagram indicating the direction of the K-field between the plates. Calculate the magnitude of the electric field strength. If a 3.4 mug charged particle is suspended between the plates, calculate the sign and magnitude of the charge. Two charges +25 muC and -50 muC are 28 cm apart. Point P is 24 cm above the midpoint of the line joining the charges. Calculate the electric potential at P. Calculate the magnitude and direction of the electric field strength at P.

Explanation / Answer

E = (550-(-225))/(0.45x10^-2) = 1.72 x10^5 V/m

F = qE = mg

q = mg/E = 3.4 x10^-6*9.8/1.72 x10^5

=1.94 x10^-10 C

it is negative charged

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