Capacitor Voltage: Calculate It From Charge and Capacitance

Calculate the voltage across a capacitor by dividing its charge by its capacitance: V = Q / C. Enter charge in coulombs and capacitance in farads to get volts; this calculator uses a practical microcoulomb-and-microfarad example and also reports the equivalent charge.

Formula

The basic relationship for a capacitor is V = Q / C. Here, V is the voltage across the capacitor in volts (V), Q is the charge stored on either plate in coulombs (C), and C is capacitance in farads (F). The calculator accepts charge in microcoulombs (µC) and capacitance in microfarads (µF); because both use the same millionth-unit scale, dividing their displayed values gives volts directly.

For example, 1 µC divided by 1 µF equals 1 V. In base SI units, the same calculation is (1 × 10⁻⁶ C) / (1 × 10⁻⁶ F) = 1 V. A farad is a coulomb per volt, so the units reduce to volts.

Worked Example

Using the calculator defaults, the charge is 10 µC and the capacitance is 2 µF.

1. Apply V = Q / C.
2. Substitute the matching micro-units: V = 10 µC / 2 µF.
3. Divide: V = 5 V.
4. Convert the charge for the helper output: 10 µC × 0.000001 = 0.000010 C.

The calculated voltage is 5.00 V, and the charge is 0.000010 C. The result assumes the entered charge and capacitance describe the same capacitor state.

What Voltage Across a Capacitor Means

A capacitor stores separated electric charge on two conductive plates, with an insulating material between them. The voltage is the electrical potential difference between the plates. For an ideal, linear capacitor, more stored charge means proportionally more voltage; increasing capacitance means the same charge produces less voltage.

The relationship can also be rearranged: Q = C × V gives charge when voltage and capacitance are known, while C = Q / V gives capacitance when charge and voltage are known. These are three forms of the same definition, not separate physical laws. The simple formula applies directly when capacitance can be treated as constant.

Using the Calculator and Choosing Units

Enter the magnitude of the charge stored on one plate and the capacitance value. The calculator returns the voltage in volts. The charge on the other plate has equal magnitude and opposite sign; do not add the two plate charges together as Q in this formula.

• Use µC with µF as shown, or convert both quantities to coulombs and farads before calculating.
• For nanocoulombs and nanofarads, their shared 10⁻⁹ scale also cancels. Likewise, matching prefixes cancel when used for both quantities.
• If the prefixes differ, convert carefully. For instance, 10 µC is 0.000010 C, while 2 mF is 0.002 F; 10 µC / 2 mF = 0.005 V.
• Keep the units consistent. A value entered as 2 F is one million times larger than 2 µF.

Capacitance is commonly labeled in farads and submultiples such as microfarads (µF), nanofarads (nF), and picofarads (pF). Charge may be expressed in coulombs or smaller units. Voltage is measured in volts; there is no separate US customary unit for electrical voltage.

When the Simple Formula Needs More Context

For a capacitor whose capacitance changes with voltage, Q/C may not describe the full relationship over a wide range. In that case, the local differential capacitance is defined as dQ/dV, and the charge-voltage curve or component data is needed. The basic calculator is intended for a stated, effectively constant capacitance.

In a circuit, voltage can also change with time as a capacitor charges or discharges. The equation V = Q/C still relates instantaneous charge and voltage for an ideal fixed capacitor, but it does not by itself predict how quickly the voltage changes. That requires circuit information such as resistance, the source, and the initial condition.

A capacitor connected to a voltage source is often described using Q = C × V. This tells you the charge associated with the stated capacitance and voltage; it does not mean the voltage can be inferred from capacitance alone. To calculate voltage using Q/C, you need both charge and capacitance.

Common Errors and Limits

• Mixing prefixes: 10 µC / 2 F is not 5 V; it is 0.000005 V. Convert units or ensure matching prefixes.
• Confusing µF with F: 1 µF equals 0.000001 F. This conversion can change an answer by a factor of one million.
• Using total charge on both plates: Q denotes the magnitude on one plate, not the sum of the absolute charges on both plates.
• Assuming a component value is exact: Real capacitors have tolerances, and capacitance may vary with conditions and component type. A computed answer is only as precise as the input values.
• Rounding too early: Retain additional digits during conversions and round the final voltage to a precision supported by the measurements.

This calculation does not determine whether a capacitor or circuit can safely withstand the result. Check the component’s specifications and the circuit design; stored charge and voltage can remain after power is removed. Use appropriate procedures and equipment when working on electrical circuits, and do not assume a capacitor is discharged.

Frequently Asked Questions

What is the formula for voltage across a capacitor?
For a fixed-capacitance capacitor, voltage equals charge divided by capacitance: V = Q/C. Use coulombs and farads, or matching prefixes such as microcoulombs and microfarads.

How do you calculate capacitor voltage from capacitance and charge?
Divide the charge on one plate by the capacitance. For example, 10 µC / 2 µF = 5 V.

Can you calculate voltage if you know only the capacitor’s capacitance?
No. You also need charge, or other circuit information that lets you determine the voltage. Capacitance alone does not specify the capacitor’s present state.

Does capacitor voltage have polarity?
Voltage is a difference between two points, so its sign depends on which plate or terminal is chosen as the reference. The formula gives a magnitude when Q is entered as a positive charge magnitude; signed circuit calculations must use a consistent polarity convention.

Why does capacitor voltage change over time?
As charge moves onto or off the plates, Q changes, and so does V for a fixed capacitance. The rate of change depends on the surrounding circuit, not on V = Q/C alone.

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in q | Charge | µC | 10 | 0
in c | Capacitance | µF | 2 | 0
out v | Voltage across capacitor | V | q / c | 2
out charge | Charge | C | q * 0.000001 | 6