A microscopic droplet of ink in an ink-jet printer is deflected onto the paper by a force of 2.34×10-4 N. If the field causing the deflection is 6.62 kN/C, what is the charge on the droplet?
What equation do I use to solve and What does each value mean in the equation? thnks
To solve this problem, you can use the equation:
F = qE
Where:
- F is the force acting on the droplet in Newtons (N)
- q is the charge on the droplet in Coulombs (C)
- E is the electric field strength in Newtons per Coulomb (N/C)
In this equation, the force (F) is given as 2.34×10-4 N, and the electric field strength (E) is given as 6.62 kN/C. The goal is to find the charge (q) on the droplet.
You need to rearrange the equation to solve for q:
q = F / E
Now you can substitute the given values into the equation:
q = 2.34×10-4 N / 6.62 kN/C
Make sure to convert kN to N by multiplying by 1000:
q = 2.34×10-4 N / (6.62 × 1000 N/C)
Simplifying the expression:
q = 2.34×10-4 N / 6620 N/C
Finally, calculate the value of q:
q ≈ 3.53 × 10-8 C
Therefore, the charge on the droplet is approximately 3.53 × 10-8 Coulombs.
To solve this problem, you can use the formula:
F = q * E
Where:
F is the force acting on the droplet (2.34×10^(-4) N),
q is the charge on the droplet (what we're trying to find),
and E is the electric field strength (6.62 kN/C).
Rearranging the equation:
q = F / E
Now we can substitute the given values into the equation:
q = (2.34×10^(-4) N) / (6.62 kN/C)
Before we proceed, let's convert the units to be consistent. Recall that 1 kN = 1000 N and 1 C = 1 C/1 C = 1.
q = (2.34×10^(-4) N) / (6.62 × 1000 N/C)
q = 2.34×10^(-4) / 6620
q ≈ 3.53 × 10^(-8) C
Therefore, the charge on the droplet is approximately 3.53 × 10^(-8) Coulombs.
F=Eq
q=F/E
q=2.34^-4/6620